Chapter 1 Real Numbers (Additional Questions)

The correct option is (C).

According to Euclid's Division Lemma, for given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying $a = bq + r$, where the remainder $r$ must satisfy the condition $0 \leq r < b$.

This means the remainder $r$ can be zero (when $a$ is a multiple of $b$) and must be strictly less than the divisor $b$.

Therefore, the condition is $0 \leq r < b$.

The correct option is (C).

The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number itself or can be represented as the product of prime numbers, and that, apart from their order, this representation is unique.

Let's examine each statement:

(A) Every composite number can be expressed as a product of primes. This is TRUE and is a fundamental part of the theorem.

(B) The prime factorization of a composite number is unique. This is also TRUE. The theorem guarantees the uniqueness of the set of prime factors and their exponents.

(C) The order of prime factors is unique. This is FALSE. The theorem explicitly states that the uniqueness is "apart from their order". For example, the prime factorization of 12 can be written as $2 \times 2 \times 3$ or $2 \times 3 \times 2$ or $3 \times 2 \times 2$. The prime factors ($2, 2, 3$) are unique, but their sequence is not necessarily unique.

(D) It relates to the prime factorization of integers greater than 1. This is TRUE, as the theorem applies to composite numbers (which are $>1$) and implicitly to prime numbers themselves (e.g., the prime factorization of 7 is just 7).

Therefore, the false statement is (C).

The correct option is (B).

To Find: The HCF of 135 and 225.

Solution:

We use Euclid's Division Algorithm to find the HCF of 135 and 225.

Since $225 > 135$, we apply the division lemma to 225 and 135.

The remainder is $90 \neq 0$. We now apply the division lemma to the new divisor 135 and the new remainder 90.

The remainder is $45 \neq 0$. We again apply the division lemma to the new divisor 90 and the new remainder 45.

The remainder is now 0. The divisor at this stage is 45.

The HCF of the given numbers is the divisor at the stage where the remainder is zero.

Therefore, the HCF of 135 and 225 is 45.

The correct option is (B).

To Find: The prime factorization of 144.

Solution:

We find the prime factorization of 144 using the division method:

$\begin{array}{c|cc} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

From the prime factorization, we can write 144 as the product of its prime factors:

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

In exponential form, this is:

$144 = 2^4 \times 3^2$

Comparing this result with the given options, we find that option (B) matches our result.

The correct option is (C).

Explanation:

Two integers $a$ and $b$ are said to be co-prime or relatively prime if their greatest common divisor (GCD) or highest common factor (HCF) is 1.

This means that the only positive integer that divides both $a$ and $b$ is 1.

Given that HCF$(a, b) = 1$, by definition, the numbers $a$ and $b$ are co-prime numbers.

Let's look at other options:

(A) Prime numbers: If $a$ and $b$ are prime numbers, their HCF can be 1 (e.g., HCF(3, 5) = 1). However, two numbers can be co-prime without being prime (e.g., HCF(4, 9) = 1, but 4 and 9 are composite). So, HCF(a,b)=1 does not imply that $a$ and $b$ are prime numbers.

(B) Composite numbers: If $a$ and $b$ are composite numbers, their HCF can be 1 (e.g., HCF(4, 9) = 1). But it can also be greater than 1 (e.g., HCF(6, 8) = 2). Also, one or both numbers can be prime or 1 and still be co-prime (e.g., HCF(2, 9) = 1, HCF(1, 5) = 1). So, HCF(a,b)=1 does not imply that $a$ and $b$ are composite numbers.

(D) Twin primes: Twin primes are pairs of prime numbers that differ by 2 (e.g., 3 and 5). Twin primes are always co-prime (since they are prime and their difference is small, they don't share common factors other than 1), but co-prime numbers are not necessarily twin primes (e.g., 4 and 9 are co-prime but not twin primes).

The definition that directly matches the condition HCF$(a, b) = 1$ is co-prime numbers.

The correct option is (B).

To Find: The LCM of 12 and 18.

Solution:

We can find the LCM of 12 and 18 using the prime factorization method or the division method. Let's use the division method as per the specified format:

$\begin{array}{c|cc} 2 & 12 \;, & 18 \\ \hline 2 & 6 \; , & 9 \\ \hline 3 & 3 \; , & 9 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$

The LCM is the product of the prime factors obtained in the division process.

LCM$(12, 18) = 2 \times 2 \times 3 \times 3$

LCM$(12, 18) = 4 \times 9$

LCM$(12, 18) = 36$

Alternatively, using prime factorization:

Prime factors of 12: $12 = 2^2 \times 3^1$

Prime factors of 18: $18 = 2^1 \times 3^2$

The LCM is found by taking the highest power of all prime factors present in the factorizations.

Prime factors are 2 and 3.

Highest power of 2 is $2^2$.

Highest power of 3 is $3^2$.

LCM$(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36$

Both methods give the same result.

The LCM of 12 and 18 is 36.

The correct option is (C).

Explanation:

For any two positive integers, the product of their HCF (Highest Common Factor) and LCM (Least Common Multiple) is equal to the product of the numbers themselves.

This property is a fundamental theorem in number theory.

Mathematically, for two positive integers $a$ and $b$, the relationship is given by:

$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

Let's verify this with an example. Consider $a = 12$ and $b = 18$.

HCF$(12, 18) = 6$ (as calculated in a previous question).

LCM$(12, 18) = 36$ (as calculated in a previous question).

Product of HCF and LCM: $6 \times 36 = 216$.

Product of the numbers: $12 \times 18 = 216$.

Since $216 = 216$, the relationship $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ holds true.

The correct option is (B).

To Find: LCM$(26, 91)$.

Solution:

We know the relationship between the HCF and LCM of two positive integers $a$ and $b$:

$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

Given $a = 26$, $b = 91$, and HCF$(26, 91) = 13$.

Substitute the given values into the formula:

To find LCM$(26, 91)$, we rearrange the equation:

$\text{LCM}(26, 91) = \frac{26 \times 91}{13}$

Now, we can calculate the value:

$\text{LCM}(26, 91) = \frac{\cancel{26}^{2} \times 91}{\cancel{13}_{1}}$

$\text{LCM}(26, 91) = 2 \times 91$

$\text{LCM}(26, 91) = 182$

Comparing this result with the given options, we find that option (B) matches the expression we derived before calculating the final value.

The expression for LCM$(26, 91)$ is $\frac{26 \times 91}{13}$.

The correct option is (C).

Explanation:

A number is considered rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Rational numbers include terminating or repeating decimals.

A number is considered irrational if it cannot be expressed in the form $\frac{p}{q}$. Irrational numbers have non-terminating and non-repeating decimal expansions.

Let's evaluate each option:

(A) $\sqrt{16}$: $\sqrt{16} = 4$. Since $4 = \frac{4}{1}$, it can be expressed as a fraction of two integers. Thus, $\sqrt{16}$ is a rational number.

(B) $\sqrt{9}$: $\sqrt{9} = 3$. Since $3 = \frac{3}{1}$, it can be expressed as a fraction of two integers. Thus, $\sqrt{9}$ is a rational number.

(C) $\sqrt{20}$: $20$ is not a perfect square. $\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$. Since $\sqrt{5}$ is an irrational number (its decimal expansion is non-terminating and non-repeating), the product of a non-zero rational number (2) and an irrational number ($\sqrt{5}$) is irrational. Thus, $\sqrt{20}$ is an irrational number.

(D) $\sqrt{49}$: $\sqrt{49} = 7$. Since $7 = \frac{7}{1}$, it can be expressed as a fraction of two integers. Thus, $\sqrt{49}$ is a rational number.

Therefore, the only irrational number among the given options is $\sqrt{20}$.

The correct option is (A).

Explanation:

Rational numbers are numbers that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

The decimal expansion of a rational number has two possible forms:

1. Terminating: The decimal expansion ends after a finite number of digits (e.g., $\frac{1}{4} = 0.25$, $\frac{3}{8} = 0.375$).

2. Non-terminating recurring (or repeating): The decimal expansion does not end, but a block of digits repeats infinitely (e.g., $\frac{1}{3} = 0.333... = 0.\overline{3}$, $\frac{1}{7} = 0.142857142857... = 0.\overline{142857}$).

Irrational numbers, on the other hand, have decimal expansions that are non-terminating and non-recurring (e.g., $\sqrt{2}$, $\pi$).

Therefore, the decimal expansion of a rational number is either terminating or non-terminating recurring.

The correct option is (A).

To Determine: The nature of the decimal expansion of $\frac{17}{8}$.

Solution:

A rational number $\frac{p}{q}$ (where $p$ and $q$ are integers, $q \neq 0$, and HCF$(p, q) = 1$) has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.

The given number is $\frac{17}{8}$. Here, $p = 17$ and $q = 8$.

First, we check if HCF$(17, 8) = 1$. Since 17 is a prime number and 8 is not a multiple of 17, their HCF is indeed 1.

Now, we find the prime factorization of the denominator, $q = 8$.

$\begin{array}{c|cc} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

The prime factorization of 8 is $2 \times 2 \times 2 = 2^3$.

We can write this in the form $2^m \times 5^n$ as $2^3 \times 5^0$ (since $5^0 = 1$).

Since the prime factors of the denominator (8) are only 2s (and no other primes), the decimal expansion of $\frac{17}{8}$ is terminating.

Let's perform the division to see the terminating decimal expansion:

$\frac{17}{8} = 2.125$ which terminates.

The correct option is (B).

To Determine: The nature of the decimal expansion of $\frac{64}{455}$.

Solution:

We are given the rational number $\frac{64}{455}$. Let $p = 64$ and $q = 455$.

To determine the nature of the decimal expansion without long division, we examine the prime factors of the denominator $q$, after ensuring the fraction is in its simplest form.

First, let's find the prime factors of the numerator $p = 64$ and the denominator $q = 455$ to check if they have any common factors.

Prime factorization of 64:

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $64 = 2^6$.

Prime factorization of 455:

$\begin{array}{c|cc} 5 & 455 \\ \hline 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

So, $455 = 5 \times 7 \times 13$.

The prime factors of the numerator (64) are only 2s. The prime factors of the denominator (455) are 5, 7, and 13. There are no common factors between the numerator and the denominator other than 1. Thus, the fraction $\frac{64}{455}$ is already in its simplest form.

Now, we examine the prime factors of the denominator $q = 455$. The prime factors are 5, 7, and 13.

A rational number $\frac{p}{q}$ (in simplest form) has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.

In this case, the prime factorization of the denominator 455 includes the prime factors 7 and 13, which are not 2 or 5.

Therefore, the decimal expansion of $\frac{64}{455}$ is non-terminating recurring.

The correct option is (A).

Explanation:

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

A key theorem regarding the decimal expansion of rational numbers states the following:

A rational number $\frac{p}{q}$, where $p$ and $q$ are co-prime integers (i.e., HCF$(p, q) = 1$) and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.

If the prime factorization of the denominator $q$ contains any prime factor other than 2 or 5, then the decimal expansion of $\frac{p}{q}$ will be non-terminating but recurring.

Therefore, for a rational number to have a terminating decimal expansion, the prime factors of its denominator must only be 2s and/or 5s. This is represented by the form $2^m \times 5^n$, where $m \geq 0$ and $n \geq 0$.

The correct option is (B).

Explanation:

A rational number is any number that can be expressed as a fraction $\frac{p}{q}$ of two integers, where $p$ is an integer and $q$ is a non-zero integer. In decimal form, rational numbers are either terminating or non-terminating recurring.

An irrational number is a number that cannot be expressed as a simple fraction. Its decimal expansion is non-terminating and non-recurring.

Let's examine each option:

(A) $0.121212\dots$: This is a non-terminating, repeating decimal ($0.\overline{12}$). Repeating decimals represent rational numbers. For example, $0.\overline{12} = \frac{12}{99} = \frac{4}{33}$.

(B) $0.1234567891011\dots$: This decimal expansion is non-terminating, and the digits do not repeat in a fixed pattern (the sequence of digits seems to be formed by concatenating consecutive integers: 1, 2, 3, ..., 9, 10, 11, ...). This is characteristic of an irrational number.

(C) $0.\overline{5}$: This is a non-terminating, repeating decimal ($0.555\dots$). Repeating decimals represent rational numbers. For example, $0.\overline{5} = \frac{5}{9}$.

(D) $\frac{3}{7}$: This is already in the form $\frac{p}{q}$, where $p=3$ and $q=7$ are integers and $q \neq 0$. By definition, this is a rational number. Its decimal expansion is $0.\overline{428571}$, which is non-terminating and recurring.

Therefore, the number that is NOT a rational number is $0.1234567891011\dots$, which is an irrational number.

The correct option is (A).

Explanation:

Assertion (A): $\sqrt{3}$ is an irrational number.

This statement is True. It is a well-known fact in number theory that the square root of any non-perfect square positive integer is an irrational number. Since 3 is not a perfect square, $\sqrt{3}$ is irrational.

Reason (R): A number is irrational if it cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

This statement is also True. This is the standard definition of an irrational number. A rational number is defined as a number that can be expressed in the form $\frac{p}{q}$, and an irrational number is one that cannot be expressed in this form.

Now let's check if Reason (R) is the correct explanation for Assertion (A).

To prove that $\sqrt{3}$ is an irrational number, the proof relies directly on the definition of an irrational number as stated in Reason (R). The typical proof involves assuming, for contradiction, that $\sqrt{3}$ *can* be expressed in the form $\frac{p}{q}$ (in its simplest form) and then showing that this assumption leads to a contradiction, thus proving that $\sqrt{3}$ *cannot* be expressed in that form and is therefore irrational according to the definition in (R).

Therefore, Reason (R) provides the fundamental definition that is used to determine whether a number like $\sqrt{3}$ is irrational. It explains *why* $\sqrt{3}$ is called irrational.

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

The correct option is (A).

Explanation:

Let's evaluate the Assertion (A).

The given number is $5 \times 7 \times 11 + 11$.

We can factor out 11 from both terms:

$5 \times 7 \times 11 + 11 = 11 \times (5 \times 7 + 1)$

$ = 11 \times (35 + 1)$

$ = 11 \times 36$

The number can be expressed as the product of two integers, 11 and 36, both of which are greater than 1.

A composite number is defined as a positive integer greater than 1 that is not prime, i.e., it has at least one divisor other than 1 and itself.

Since $11 \times 36 = 396$, the number 396 is divisible by 11 and 36 (among other factors like 2, 3, 4, etc.), which are factors other than 1 and 396. Thus, 396 is a composite number.

Therefore, Assertion (A) is True.

Now, let's evaluate the Reason (R).

Reason (R) states: A composite number is a number that has factors other than 1 and itself.

This is the standard definition of a composite number. A positive integer greater than 1 is composite if it has more than two distinct positive divisors (1 and the number itself). These additional divisors are the "factors other than 1 and itself".

Therefore, Reason (R) is True.

Finally, let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) claims that $5 \times 7 \times 11 + 11$ is composite. Our calculation showed that this number can be factored into $11 \times 36$, meaning it has factors 11 and 36 (which are neither 1 nor the number itself, 396). This conclusion that the number is composite is based directly on the definition of a composite number provided in Reason (R).

Therefore, Reason (R) correctly explains why Assertion (A) is true.

Both A and R are true, and R is the correct explanation of A.

The correct option is (A).

Explanation:

Let's match the number types with their examples:

(i) Rational Number: A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Also includes terminating or non-terminating recurring decimals.

Example (c) $\frac{2}{5}$ is in the form $\frac{p}{q}$, so it is a rational number.

Match: (i)-(c)

(ii) Irrational Number: A number that cannot be expressed in the form $\frac{p}{q}$. Its decimal expansion is non-terminating and non-recurring.

Example (b) $0.101001000\dots$ is a non-terminating, non-recurring decimal. The number of zeros between the ones increases, so there is no repeating block of digits. Thus, it is an irrational number.

Match: (ii)-(b)

(iii) Prime Number: A natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.

Example (a) 17 is a prime number as its only positive divisors are 1 and 17.

Match: (iii)-(a)

(iv) Composite Number: A positive integer greater than 1 that has at least one positive divisor other than 1 and itself. In other words, it can be formed by multiplying two smaller positive integers.

Example (d) 12 is a composite number as it has divisors other than 1 and 12 (e.g., 2, 3, 4, 6). $12 = 3 \times 4 = 2 \times 6$.

Match: (iv)-(d)

Combining the matches: (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d).

This combination corresponds to option (A).

The correct option is (C).

Explanation:

The sweet shop owner wants to pack laddoos of two types (implied total number or weight units being 144 and 180) into boxes such that each box contains the same number of laddoos, and this number is the maximum possible.

This means he needs to find the largest number that can divide both 144 and 180 without leaving a remainder.

The largest number that divides two or more integers is known as their Highest Common Factor (HCF) or Greatest Common Divisor (GCD).

Let's consider the options:

(A) LCM (Least Common Multiple) is used to find the smallest common multiple of two or more numbers, which is not required in this scenario.

(B) Prime Factorization is a method used to find the prime factors of a number, which can then be used to calculate HCF or LCM. It is a tool, not the core concept being applied to solve the problem.

(C) HCF (Highest Common Factor) is the largest number that divides two or more numbers. Finding the HCF of 144 and 180 will give the maximum number of laddoos that can be packed in each box.

(D) Euclid's Division Lemma is an algorithm to find the HCF of two numbers. Like prime factorization, it is a method or tool to calculate the HCF, not the underlying concept of finding the maximum common divisor.

Therefore, the mathematical concept needed is the HCF.

This question requires us to refer to the scenario provided in Question 18. Since the details of the sweet shop scenario from Question 18 are not provided here, we will assume typical quantities for such a problem that would lead to one of the given options (12, 18, 36, 72) as the answer. The question asks for the maximum number of laddoos that can be packed in each box, which means we need to find the greatest common divisor (GCD) of the number of laddoos of each type mentioned in Question 18.

Let's assume, for example, that the sweet shop has 324 Bundi laddoos and 396 Motichoor laddoos. We need to find the maximum number of laddoos per box such that each box contains only one type of laddoo, and the number of laddoos is the same in every box. This number is the GCD of 324 and 396.

We can find the GCD using prime factorization:

Prime factorization of 324:

$324 = 2 \times 162$

$324 = 2 \times 2 \times 81$

$324 = 2^2 \times 3^4$

Prime factorization of 396:

$396 = 2 \times 198$

$396 = 2 \times 2 \times 99$

$396 = 2^2 \times 3 \times 33$

$396 = 2^2 \times 3^2 \times 11$

The greatest common divisor is the product of the lowest powers of the common prime factors.

The common prime factors are 2 and 3.

The lowest power of 2 is $2^2$.

The lowest power of 3 is $3^2$.

GCD$(324, 396) = 2^2 \times 3^2 = 4 \times 9 = 36$.

Thus, based on the assumption of 324 and 396 laddoos, the maximum number of laddoos that can be packed in each box is 36.

This result (36) matches option (C). Therefore, it is highly probable that the quantities of laddoos in Question 18 were numbers whose GCD is 36.

The correct option is (C) 36.

We are given the fraction $\frac{23}{2^2 \times 5^1}$.

A rational number in its simplest form has a terminating decimal expansion if and only if the prime factorization of the denominator contains only powers of 2 and 5.

In this case, the fraction is $\frac{23}{2^2 \times 5^1}$. The numerator 23 is a prime number, and the denominator is $2^2 \times 5^1$. Since the prime factors in the denominator are only 2 and 5, the decimal expansion of this fraction will terminate.

To find the number of decimal places after which the expansion terminates, we look at the powers of 2 and 5 in the prime factorization of the denominator. The number of decimal places is equal to the maximum of the powers of 2 and 5.

The denominator is $2^2 \times 5^1$.

The power of 2 is 2.

The power of 5 is 1.

The maximum of these powers is $\text{max}(2, 1) = 2$.

Therefore, the decimal expansion of $\frac{23}{2^2 \times 5^1}$ will terminate after 2 decimal places.

We can also calculate the decimal expansion to verify:

$\frac{23}{2^2 \times 5^1} = \frac{23}{4 \times 5} = \frac{23}{20}$

To convert this to a decimal, we can divide 23 by 20 or multiply the numerator and denominator by 5 to make the denominator a power of 10:

$\frac{23}{20} = \frac{23 \times 5}{20 \times 5} = \frac{115}{100} = 1.15$

The decimal expansion is 1.15, which terminates after 2 decimal places.

The correct option is (B) 2.

We are asked to identify the number that is not a real number among the given options.

A real number is any number that can be found on the number line. Real numbers include rational numbers (integers, fractions) and irrational numbers (like $\sqrt{2}$ or $\pi$). The set of real numbers is denoted by $\mathbb{R}$.

Let's examine each option:

(A) $5$: This is an integer, which is a rational number ($5 = \frac{5}{1}$). Rational numbers are real numbers.

(B) $\sqrt{7}$: This is the square root of a number that is not a perfect square. It is an irrational number. Irrational numbers are real numbers.

(C) $\pi$: This is a mathematical constant, approximately 3.14159.... It is an irrational number. Irrational numbers are real numbers.

(D) $\sqrt{-4}$: This is the square root of a negative number. The square root of a negative number is an imaginary number, denoted by $i = \sqrt{-1}$.

We can write $\sqrt{-4} = \sqrt{4 \times -1} = \sqrt{4} \times \sqrt{-1} = 2 \times i = 2i$.

Imaginary numbers and complex numbers (numbers of the form $a + bi$, where $a$ and $b$ are real numbers and $b \neq 0$) are not real numbers.

Therefore, $\sqrt{-4}$ is not a real number.

The correct option is (D) $\sqrt{-4}$.

We are asked to find the Highest Common Factor (HCF) of two consecutive integers.

Consecutive integers are integers that follow each other in order. They differ by 1. Examples include (2, 3), (10, 11), (-5, -4), etc. Let the two consecutive positive integers be $n$ and $n+1$, where $n$ is a positive integer.

The HCF of two numbers is the largest positive integer that divides both numbers without leaving a remainder.

Let $d$ be a common divisor of $n$ and $n+1$.

If $d$ divides $n$, then $n = k \times d$ for some integer $k$.

If $d$ also divides $n+1$, then $n+1 = m \times d$ for some integer $m$.

The difference between the two consecutive integers is $(n+1) - n = 1$.

If a number $d$ divides two integers, it must also divide their difference.

So, $d$ must divide $(n+1) - n$, which means $d$ must divide 1.

The only positive integer that divides 1 is 1 itself.

Therefore, the only positive common divisor of two consecutive integers $n$ and $n+1$ is 1.

Since 1 is the only positive common divisor, it is also the highest common divisor (HCF).

Let's take an example: Find the HCF of 7 and 8.

Factors of 7 are 1, 7.

Factors of 8 are 1, 2, 4, 8.

The common factor is only 1.

HCF$(7, 8) = 1$.

Another example: Find the HCF of 15 and 16.

Factors of 15 are 1, 3, 5, 15.

Factors of 16 are 1, 2, 4, 8, 16.

The common factor is only 1.

HCF$(15, 16) = 1$.

In all cases, the HCF of two consecutive integers is 1.

The correct option is (C) 1.

We are given that $p$ is a prime number and $p$ divides $a^2$, where $a$ is a positive integer.

We need to determine which of the given options $p$ must divide.

This question is based on a fundamental property of prime numbers, often referred to as Euclid's Lemma or a consequence of the Unique Prime Factorization Theorem (Fundamental Theorem of Arithmetic).

Euclid's Lemma states that if a prime number $p$ divides the product of two integers, say $xy$, then $p$ must divide at least one of the integers $x$ or $y$. That is, if $p | xy$, then $p | x$ or $p | y$.

In our case, we are given that $p$ divides $a^2$. We can write $a^2$ as the product $a \times a$.

So, we have $p | (a \times a)$.

Applying Euclid's Lemma, if $p$ divides the product $a \times a$, then $p$ must divide the first factor ($a$) or the second factor ($a$).

Since both factors are the same ($a$), the lemma tells us that $p$ must divide $a$.

Let's consider the prime factorization of $a$. Let $a = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$, where $p_1, p_2, \dots, p_k$ are distinct prime factors of $a$ and $e_i \ge 1$.

Then $a^2 = (p_1^{e_1} p_2^{e_2} \dots p_k^{e_k})^2 = p_1^{2e_1} p_2^{2e_2} \dots p_k^{2e_k}$.

We are given that $p$ divides $a^2$. This means that $p$ is a prime factor of $a^2$.

According to the Unique Prime Factorization Theorem, the prime factorization of $a^2$ is unique. The prime factors of $a^2$ are $p_1, p_2, \dots, p_k$.

Since $p$ is a prime factor of $a^2$, $p$ must be one of the primes $p_1, p_2, \dots, p_k$.

If $p$ is one of the primes $p_i$ in the prime factorization of $a$, then $p$ must divide $a$.

Let's check the options:

(A) $a$: Our conclusion is that $p$ divides $a$.

(B) $\sqrt{a}$: $\sqrt{a}$ may not be an integer, so the concept of $p$ dividing $\sqrt{a}$ is not applicable in the context of integer divisibility.

(C) $a^3$: If $p$ divides $a$, then $a = pk$ for some integer $k$. Then $a^3 = (pk)^3 = p^3 k^3 = p \times (p^2 k^3)$. So $p$ divides $a^3$. However, this is a consequence of $p|a$, which is the primary result.

(D) $2a$: If $p$ divides $a$, then $a = pk$. Then $2a = 2(pk) = p \times (2k)$. So $p$ divides $2a$. This is also a consequence of $p|a$.

The most direct and fundamental conclusion from $p | a^2$ (where $p$ is prime) is that $p$ must divide $a$. Options (C) and (D) are true if $p|a$, but $p|a$ is what is directly implied by $p|a^2$.

The correct option is (A) $a$.

We are given the number $0.1212212221\dots$ and asked to classify it.

Numbers can be classified as rational or irrational.

• A rational number is a number that can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The decimal expansion of a rational number is either terminating or non-terminating repeating.
• An irrational number is a number that cannot be expressed as a fraction $\frac{p}{q}$. The decimal expansion of an irrational number is non-terminating and non-repeating.

Let's examine the given number: $0.1212212221\dots$

The digits after the decimal point are $12, 122, 1222, 12222, \dots$. The pattern involves a '1' followed by an increasing number of '2's, then another '1', and so on.

The ellipsis ($\dots$) indicates that the decimal expansion continues indefinitely, so it is a non-terminating decimal.

There is no block of digits that repeats itself periodically. The sequence of digits ($1212212221\dots$) does not show a repeating pattern. For example, the block '12' is followed by '122', then '1222', which are different.

Thus, it is a non-repeating decimal.

Since the decimal expansion is both non-terminating and non-repeating, the number $0.1212212221\dots$ is an irrational number.

Let's consider the given options:

(A) Rational: Incorrect, as the decimal is non-repeating.

(B) Irrational: Correct, as the decimal is non-terminating and non-repeating.

(C) Integer: Incorrect, integers are whole numbers and their negatives ($..., -1, 0, 1, ...$). The given number is between 0 and 1.

(D) Whole: Incorrect, whole numbers are non-negative integers ($0, 1, 2, ...$). The given number is between 0 and 1 and is not an integer.

The correct option is (B) Irrational.

We are asked to find the Highest Common Factor (HCF) of two prime numbers.

Recall the definition of a prime number: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Recall the definition of HCF: The Highest Common Factor (HCF) of two or more integers is the largest positive integer that divides each of the integers without leaving a remainder.

Let the two prime numbers be $p_1$ and $p_2$.

Case 1: The two prime numbers are distinct, i.e., $p_1 \neq p_2$.

The positive divisors of $p_1$ are 1 and $p_1$.

The positive divisors of $p_2$ are 1 and $p_2$.

Since $p_1$ and $p_2$ are distinct prime numbers, they do not share any prime factors other than 1. Therefore, the only common positive divisor is 1.

The HCF$(p_1, p_2) = 1$ when $p_1 \neq p_2$.

Case 2: The two prime numbers are the same, i.e., $p_1 = p_2 = p$, where $p$ is a prime number.

The positive divisors of $p$ are 1 and $p$.

The two numbers are $p$ and $p$. The common positive divisors are 1 and $p$.

The largest common divisor is $p$.

The HCF$(p, p) = p$ when the primes are the same.

Looking at the options provided:

(A) The product of the numbers

(B) 1

(C) The smaller prime

(D) 0

Option (A) is incorrect (e.g., HCF of 2 and 3 is 1, product is 6).

Option (D) is incorrect (HCF is always a positive integer for non-zero numbers).

Option (C) "The smaller prime" would imply HCF$(2, 3) = 2$, which is false (it's 1). This option only makes sense if the primes are different, but it gives the wrong value.

Option (B) "1" is the correct HCF for any two distinct prime numbers. This is a fundamental property: distinct primes are always co-prime.

While the HCF of two identical primes $p$ is $p$, the options suggest that the question likely refers to the more general case or the property of distinct primes. Option (B) is the only correct answer among the choices for the case of distinct primes, and it is a standard result concerning prime numbers and HCF.

Based on the typical context of such questions and the provided options, the question is best interpreted as referring to the HCF of two distinct prime numbers.

The HCF of two prime numbers (assuming distinct primes, which aligns with option (B)) is 1.

The correct option is (B) 1.

The decimal expansion of a rational number $\frac{p}{q}$ (in simplest form) is terminating if the prime factorization of the denominator $q$ contains only powers of 2 and/or 5.

The decimal expansion is non-terminating recurring if the prime factorization of the denominator $q$ contains at least one prime factor other than 2 or 5.

The question states that the decimal expansion is non-terminating recurring. This implies that the denominator must have prime factors other than 2 or 5.

Let's check the options based on this property:

(A) Only 2 or 5: Results in a terminating decimal.

(B) Primes other than 2 or 5: The presence of such primes leads to a non-terminating recurring decimal. This is the defining characteristic.

(C) Both 2, 5 and some other primes: The presence of "some other primes" (other than 2 or 5) makes the decimal non-terminating recurring. This is a specific case covered by the condition in (B).

(D) Any primes are possible: Incorrect.

For the decimal expansion to be non-terminating recurring, the denominator's prime factors must include factors from the set of primes other than 2 or 5.

The correct option is (B) Primes other than 2 or 5.

We are asked to identify the correct statement about the set of real numbers.

The set of real numbers, denoted by $\mathbb{R}$, comprises all rational and irrational numbers.

Rational numbers ($\mathbb{Q}$) are numbers that can be expressed as a fraction $\frac{p}{q}$ where $p, q \in \mathbb{Z}, q \neq 0$. Their decimal representations are either terminating or repeating.

Irrational numbers ($\mathbb{I}$) are numbers that cannot be expressed as a fraction $\frac{p}{q}$. Their decimal representations are non-terminating and non-repeating.

Every real number is either rational or irrational, and a number cannot be both rational and irrational. This means the sets of rational and irrational numbers are disjoint, and their combination forms the set of real numbers.

In set theory terms, the set of real numbers is the union of the set of rational numbers and the set of irrational numbers: $\mathbb{R} = \mathbb{Q} \cup \mathbb{I}$.

Let's examine the options:

(A) It consists only of rational numbers: Incorrect. Real numbers also include irrational numbers like $\sqrt{2}$ and $\pi$.

(B) It consists only of irrational numbers: Incorrect. Real numbers also include rational numbers like 0, 1, and $\frac{1}{2}$.

(C) It is the union of rational and irrational numbers: Correct. This statement accurately describes the composition of the set of real numbers.

(D) It is the intersection of rational and irrational numbers: Incorrect. The intersection of disjoint sets is the empty set, and the set of real numbers is not empty.

The correct statement is that the set of real numbers is the union of rational and irrational numbers.

The correct option is (C) It is the union of rational and irrational numbers.

We need to find the HCF of 4052 and 12576 using Euclid's Algorithm (also known as the Euclidean Algorithm).

Euclid's Algorithm is based on the principle that the HCF of two numbers does not change when the larger number is replaced by its difference with the smaller number. Repeated application of this principle, using the division algorithm, leads to the HCF.

We start by dividing the larger number (12576) by the smaller number (4052) and find the remainder.

Using the Division Algorithm: $a = bq + r$, where $0 \leq r < b$.

Step 1: Divide 12576 by 4052.

$12576 = 4052 \times 3 + 420$

Step 2: The remainder from Step 1 (420) becomes the new divisor, and the divisor from Step 1 (4052) becomes the new dividend. Divide 4052 by 420.

$4052 = 420 \times 9 + 272$

Step 3: The remainder from Step 2 (272) becomes the new divisor, and the divisor from Step 2 (420) becomes the new dividend. Divide 420 by 272.

$420 = 272 \times 1 + 148$

Step 4: The remainder from Step 3 (148) becomes the new divisor, and the divisor from Step 3 (272) becomes the new dividend. Divide 272 by 148.

$272 = 148 \times 1 + 124$

Step 5: The remainder from Step 4 (124) becomes the new divisor, and the divisor from Step 4 (148) becomes the new dividend. Divide 148 by 124.

$148 = 124 \times 1 + 24$

Step 6: The remainder from Step 5 (24) becomes the new divisor, and the divisor from Step 5 (124) becomes the new dividend. Divide 124 by 24.

$124 = 24 \times 5 + 4$

Step 7: The remainder from Step 6 (4) becomes the new divisor, and the divisor from Step 6 (24) becomes the new dividend. Divide 24 by 4.

$24 = 4 \times 6 + 0$

Since the remainder is now 0, the algorithm stops.

The HCF is the last non-zero remainder, which is 4 (from Step 6).

Therefore, the HCF of 4052 and 12576 is 4.

The correct option is (A) 4.

The problem asks for the maximum weight of fruit that can be put into each heap, such that each heap contains only one type of fruit and all heaps have the same weight. This means we are looking for a weight that can divide both the total weight of apples (48 kg) and the total weight of mangoes (60 kg) exactly.

The weight per heap must be a common divisor of 48 and 60, because we need to divide the total weight of each fruit into heaps of that specific weight.

Since the shopkeeper wants the maximum weight of fruit in each heap, we are looking for the largest possible common divisor of 48 and 60.

The largest common divisor of two or more numbers is called their Highest Common Factor (HCF) or Greatest Common Divisor (GCD).

Therefore, the mathematical concept relevant here is the HCF of 48 and 60.

Let's briefly look at the other options:

(A) LCM of 48 and 60: The Least Common Multiple is used to find the smallest common multiple, which is not relevant for dividing quantities into equal parts.

(B) Prime factorization of 48 and 60: Prime factorization is a method used to *calculate* the HCF or LCM, but it is not the underlying concept itself.

(D) Dividing the total weight by the number of heaps: We do not know the number of heaps beforehand, and this operation does not guarantee that the weight is a common divisor of the individual fruit weights.

The correct concept to find the maximum equal weight for heaps of different fruits is the HCF.

The correct option is (C) HCF of 48 and 60.

As determined in Question 29, the maximum weight of fruit the shopkeeper can put in each heap is the Highest Common Factor (HCF) of the weights of apples and mangoes.

The weight of apples is 48 kg.

The weight of mangoes is 60 kg.

We need to find the HCF of 48 and 60.

We can find the HCF by listing the factors of each number or by using prime factorization.

Using prime factorization:

Prime factorization of 48:

$48 = 2 \times 24$

$48 = 2 \times 2 \times 12$

$48 = 2 \times 2 \times 2 \times 6$

$48 = 2 \times 2 \times 2 \times 2 \times 3$

$48 = 2^4 \times 3^1$

Prime factorization of 60:

$60 = 2 \times 30$

$60 = 2 \times 2 \times 15$

$60 = 2 \times 2 \times 3 \times 5$

$60 = 2^2 \times 3^1 \times 5^1$

The HCF is the product of the lowest powers of the common prime factors.

The common prime factors are 2 and 3.

The lowest power of 2 is $2^2$ (since we have $2^4$ in 48 and $2^2$ in 60).

The lowest power of 3 is $3^1$ (since we have $3^1$ in 48 and $3^1$ in 60).

HCF$(48, 60) = 2^2 \times 3^1 = 4 \times 3 = 12$.

Alternatively, using Euclid's Algorithm:

Divide 60 by 48:

$60 = 48 \times 1 + 12$

Divide 48 by the remainder 12:

$48 = 12 \times 4 + 0$

The last non-zero remainder is 12. So, HCF$(48, 60) = 12$.

The maximum weight of fruit the shopkeeper can put in each heap is 12 kg.

The correct option is (B) 12 kg.

We want to determine if the number $6^n$ can end with the digit 0 for a natural number $n$, and under what condition.

A natural number ends with the digit 0 if and only if it is divisible by 10.

For a number to be divisible by 10, its prime factorization must include both 2 and 5, because $10 = 2 \times 5$.

Let's find the prime factorization of $6^n$. The base is 6, and its prime factorization is $2 \times 3$.

So, $6^n = (2 \times 3)^n = 2^n \times 3^n$.

For $6^n$ to end with the digit 0, its prime factorization must contain both a power of 2 and a power of 5.

From the prime factorization $2^n \times 3^n$, we see that $6^n$ has the prime factor 2 raised to the power of $n$. Since $n$ is a natural number, $n \ge 1$, so $2^n$ will always have 2 as a factor.

However, the prime factorization $2^n \times 3^n$ does not contain the prime factor 5, for any natural number $n$. The only prime factors are 2 and 3.

According to the Unique Prime Factorization Theorem, the prime factorization of any natural number is unique. Since the prime factorization of $6^n$ does not include 5, $6^n$ is never divisible by 5.

Since $6^n$ is not divisible by 5, it cannot be divisible by $2 \times 5 = 10$. Therefore, $6^n$ cannot end with the digit 0 for any natural number $n$.

Let's evaluate the options based on this conclusion:

(A) $n$ is even: If $n=2$, $6^2 = 36$. Does not end in 0.

(B) $n$ is odd: If $n=1$, $6^1 = 6$. If $n=3$, $6^3 = 216$. Neither ends in 0.

(C) The prime factors of the base include 2 and 5: The base is 6. The prime factors of 6 are 2 and 3. This condition is not true for the base 6.

(D) This is not possible for any natural number $n$: This statement aligns with our finding that $6^n$ never ends with the digit 0.

The inability of $6^n$ to end with the digit 0 is because its prime factorization $2^n \times 3^n$ lacks the prime factor 5.

The correct option is (D) This is not possible for any natural number $n$.

We need to find the Highest Common Factor (HCF) of the smallest prime number and the smallest composite number.

First, let's identify the smallest prime number.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The sequence of natural numbers greater than 1 is 2, 3, 4, 5, 6, ...

• 2 is divisible by 1 and 2 (prime).
• 3 is divisible by 1 and 3 (prime).
• 4 is divisible by 1, 2, and 4 (not prime).

The smallest prime number is 2.

Next, let's identify the smallest composite number.

A composite number is a natural number greater than 1 that is not prime. It has positive divisors other than 1 and itself.

From the sequence 2, 3, 4, 5, 6, ...

• 2 is prime.
• 3 is prime.
• 4 is not prime (it's composite). Its divisors are 1, 2, and 4.

The smallest composite number is 4.

Now, we need to find the HCF of 2 and 4.

The HCF of two numbers is the largest positive integer that divides both numbers without leaving a remainder.

Let's list the positive divisors:

Divisors of 2: $\{1, 2\}$

Divisors of 4: $\{1, 2, 4\}$

The common divisors of 2 and 4 are the elements in the intersection of their divisor sets: $\{1, 2\}$.

The largest number in the set of common divisors is 2.

So, HCF$(2, 4) = 2$.

Alternatively, using prime factorization:

$2 = 2^1$

$4 = 2^2$

The common prime factor is 2. The lowest power of the common prime factor is $2^1$.

HCF$(2, 4) = 2^1 = 2$.

The HCF of the smallest prime number (2) and the smallest composite number (4) is 2.

Let's check the given options:

(A) 1

(B) 2

(C) 3

(D) 4

Our calculated HCF is 2, which matches option (B).

The correct option is (B) 2.

We are asked to identify an equivalent form of Euclid's Division Lemma.

Euclid's Division Lemma (or Euclidean Division) states that for any two positive integers $a$ (dividend) and $b$ (divisor), there exist unique integers $q$ (quotient) and $r$ (remainder) such that:

where $0 \leq r < b$.

Let's examine the given options:

(A) $a = bq + r$, where $0 \leq r < b$: This is the exact standard mathematical statement of Euclid's Division Lemma for positive integers $a$ and $b$. It describes the process of dividing $a$ by $b$ to get a unique quotient $q$ and remainder $r$ that is non-negative and strictly less than the divisor $b$. This is a direct equivalent form.

(B) The division algorithm for polynomials: This is an analogous theorem that applies to polynomials. For any two polynomials $P(x)$ and $D(x) \neq 0$, there exist unique polynomials $Q(x)$ and $R(x)$ such that $P(x) = D(x)Q(x) + R(x)$, where the degree of $R(x)$ is less than the degree of $D(x)$. While structurally similar and based on the same underlying principle of division with remainder, it applies to polynomials, not integers, and is considered an analogue or extension rather than an "equivalent form" of the lemma specifically for integers.

(C) The fact that any integer can be written as $2k$ or $2k+1$: This is a specific application of Euclid's Division Lemma. If we take $b=2$, then for any integer $a$, there exist unique integers $q$ and $r$ such that $a = 2q + r$, where $0 \leq r < 2$. This means $r$ can only be 0 or 1. So, $a = 2q$ (even integer) or $a = 2q + 1$ (odd integer). This is a consequence or a special case, not an equivalent general form of the lemma.

(D) All of the above are related to or derived from the lemma: This is a true statement. (A) is the lemma itself (related/derived from itself). (B) is a related analogue. (C) is derived from the lemma. However, the question asks for an "equivalent form", and option (A) is the most direct and standard representation of that form for integers.

Among the given options, option (A) is the precise mathematical statement that defines Euclid's Division Lemma.

The correct option is (A) $a = bq + r$, where $0 \leq r < b$.

Given:

Product of two numbers = 1800

HCF (Highest Common Factor) of the two numbers = 15

To Find:

LCM (Least Common Multiple) of the two numbers.

Solution:

There is a fundamental relationship between the product of two positive integers, their HCF, and their LCM. This relationship is given by the formula:

We are given the product of the two numbers and their HCF. Let the two numbers be $a$ and $b$.

We are given $a \times b = 1800$ and HCF$(a, b) = 15$.

Using the formula (i), we can write:

$1800 = 15 \times \text{LCM}(a, b)$

To find the LCM, we need to divide the product by the HCF:

$\text{LCM}(a, b) = \frac{\text{Product of two numbers}}{\text{HCF}}$

Now, we perform the division:

$\frac{1800}{15} = \frac{1800}{15}$

We can simplify this fraction. Both 1800 and 15 are divisible by 15.

$1800 \div 15 = 120$

Let's verify: $15 \times 120 = 15 \times 12 \times 10 = 180 \times 10 = 1800$.

So, the LCM of the two numbers is 120.

The maximum weight of fruit the shopkeeper can put in each heap is 12 kg.

The correct option is (A) 120.

We are given two numbers $a$ and $b$ in their prime factorized form:

We need to find the Highest Common Factor (HCF) of $a$ and $b$.

To find the HCF of two numbers from their prime factorizations, we identify the common prime factors and take the product of these factors raised to the lowest power they appear in either factorization.

The prime factors of $a$ are 2 and 3, with powers 3 and 2, respectively ($2^3, 3^2$).

The prime factors of $b$ are 2 and 3, with powers 2 and 3, respectively ($2^2, 3^3$).

The common prime factors are 2 and 3.

For the prime factor 2, the powers are 3 (in $a$) and 2 (in $b$). The lowest power is $\min(3, 2) = 2$. So, we use $2^2$.

For the prime factor 3, the powers are 2 (in $a$) and 3 (in $b$). The lowest power is $\min(2, 3) = 2$. So, we use $3^2$.

Therefore, the HCF$(a, b)$ is the product of these lowest powers:

Let's compare this result with the given options:

(A) $2^3 \times 3^3$

(B) $2^2 \times 3^2$

(C) $2^5 \times 3^5$

(D) $2^1 \times 3^1$

Our calculated HCF is $2^2 \times 3^2$, which matches option (B).

The correct option is (B) $2^2 \times 3^2$.

We are given two numbers $a$ and $b$ in their prime factorized form:

We need to find the Least Common Multiple (LCM) of $a$ and $b$.

To find the LCM of two numbers from their prime factorizations, we identify all the prime factors that appear in either factorization and take the product of these factors raised to the highest power they appear in either factorization.

The prime factors involved in the factorizations of $a$ and $b$ are 2 and 3.

For the prime factor 2, the powers are 3 (in $a$) and 2 (in $b$). The highest power is $\max(3, 2) = 3$. So, we use $2^3$.

For the prime factor 3, the powers are 2 (in $a$) and 3 (in $b$). The highest power is $\max(2, 3) = 3$. So, we use $3^3$.

Therefore, the LCM$(a, b)$ is the product of these highest powers:

Let's compare this result with the given options:

(A) $2^3 \times 3^3$

(B) $2^2 \times 3^2$

(C) $2^5 \times 3^5$

(D) $2^1 \times 3^1$

Our calculated LCM is $2^3 \times 3^3$, which matches option (A).

The correct option is (A) $2^3 \times 3^3$.

A rational number $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ (when the fraction is in its simplest form) contains only powers of 2 and/or 5. That is, the denominator $q$ must be of the form $2^m 5^n$, where $m$ and $n$ are non-negative integers.

We need to examine each given rational number, simplify it to its lowest terms, and check the prime factors of the denominator.

(A) $\frac{125}{441}$

Prime factorization of numerator: $125 = 5 \times 5 \times 5 = 5^3$

Prime factorization of denominator: $441 = 21 \times 21 = (3 \times 7) \times (3 \times 7) = 3^2 \times 7^2$

The fraction is $\frac{5^3}{3^2 \times 7^2}$. There are no common factors between the numerator and the denominator, so it is in simplest form.

The denominator is $441 = 3^2 \times 7^2$. The prime factors of the denominator are 3 and 7. These are not 2 or 5.

Therefore, the decimal expansion of $\frac{125}{441}$ is non-terminating recurring.

(B) $\frac{77}{210}$

Prime factorization of numerator: $77 = 7 \times 11$

Prime factorization of denominator: $210 = 2 \times 105 = 2 \times 3 \times 35 = 2 \times 3 \times 5 \times 7$

The fraction is $\frac{7 \times 11}{2 \times 3 \times 5 \times 7}$. We can cancel the common factor 7:

$\frac{\cancel{7} \times 11}{2 \times 3 \times 5 \times \cancel{7}} = \frac{11}{2 \times 3 \times 5}$

This is in simplest form.

The denominator is $2 \times 3 \times 5$. The prime factors of the denominator are 2, 3, and 5. It contains the prime factor 3, which is not 2 or 5.

Therefore, the decimal expansion of $\frac{77}{210}$ is non-terminating recurring.

(C) $\frac{15}{1600}$

Prime factorization of numerator: $15 = 3 \times 5$

Prime factorization of denominator: $1600 = 16 \times 100 = 2^4 \times 10^2 = 2^4 \times (2 \times 5)^2 = 2^4 \times 2^2 \times 5^2 = 2^6 \times 5^2$

The fraction is $\frac{3 \times 5}{2^6 \times 5^2}$. We can cancel the common factor 5:

$\frac{3 \times \cancel{5}}{2^6 \times 5^{\cancel{2} 1}} = \frac{3}{2^6 \times 5^1}$

This is in simplest form.

The denominator is $2^6 \times 5^1$. The prime factors of the denominator are 2 and 5. These are the only prime factors.

Therefore, the decimal expansion of $\frac{15}{1600}$ is terminating.

(D) $\frac{29}{343}$

Prime factorization of numerator: 29 (29 is a prime number)

Prime factorization of denominator: $343 = 7 \times 49 = 7 \times 7 \times 7 = 7^3$

The fraction is $\frac{29}{7^3}$. There are no common factors between the numerator and the denominator, so it is in simplest form.

The denominator is $7^3$. The only prime factor of the denominator is 7. This is not 2 or 5.

Therefore, the decimal expansion of $\frac{29}{343}$ is non-terminating recurring.

Only the rational number $\frac{15}{1600}$ (in its simplest form $\frac{3}{2^6 \times 5^1}$) has a denominator whose prime factors are exclusively 2 and 5. Thus, its decimal expansion terminates.

The correct option is (C) $\frac{15}{1600}$.

We need to identify the pair of co-prime numbers among the given options.

Co-prime numbers (or relatively prime numbers) are two integers that have no common positive divisors other than 1. In other words, their Highest Common Factor (HCF) is 1.

Let's find the HCF for each pair of numbers:

(A) (8, 12)

Factors of 8: 1, 2, 4, 8

Factors of 12: 1, 2, 3, 4, 6, 12

Common factors: 1, 2, 4

HCF$(8, 12) = 4$. Since HCF is not 1, 8 and 12 are not co-prime.

(B) (15, 21)

Factors of 15: 1, 3, 5, 15

Factors of 21: 1, 3, 7, 21

Common factors: 1, 3

HCF$(15, 21) = 3$. Since HCF is not 1, 15 and 21 are not co-prime.

(C) (17, 23)

17 is a prime number. Its factors are 1 and 17.

23 is a prime number. Its factors are 1 and 23.

Common factors: 1

HCF$(17, 23) = 1$. Since HCF is 1, 17 and 23 are co-prime.

(D) (10, 15)

Factors of 10: 1, 2, 5, 10

Factors of 15: 1, 3, 5, 15

Common factors: 1, 5

HCF$(10, 15) = 5$. Since HCF is not 1, 10 and 15 are not co-prime.

Only the pair (17, 23) has an HCF of 1, meaning they are co-prime.

Note that any two distinct prime numbers are always co-prime.

The correct option is (C) (17, 23).

We are asked to find the product of the HCF and LCM of the numbers 6 and 20.

There is a property that states that for any two positive integers, the product of the numbers is equal to the product of their HCF and LCM.

Let the two numbers be $a$ and $b$. The property is:

In this question, the two numbers are 6 and 20.

According to the property, the product of HCF(6, 20) and LCM(6, 20) is equal to the product of the numbers 6 and 20.

Now, we calculate the product of 6 and 20:

$6 \times 20 = 120$

Therefore, the product of HCF and LCM of 6 and 20 is 120.

Let's verify this by calculating the HCF and LCM separately:

Prime factorization of 6: $2^1 \times 3^1$

Prime factorization of 20: $2^2 \times 5^1$

HCF$(6, 20) = 2^{\min(1, 2)} \times 3^{\min(1, 0)} \times 5^{\min(0, 1)} = 2^1 \times 3^0 \times 5^0 = 2 \times 1 \times 1 = 2$

LCM$(6, 20) = 2^{\max(1, 2)} \times 3^{\max(1, 0)} \times 5^{\max(0, 1)} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$

Product of HCF and LCM = $2 \times 60 = 120$.

This confirms the result obtained using the property.

The correct option is (A) 120.

We are looking for the smallest number that is divisible by both 15 and 20.

A number that is divisible by both 15 and 20 is a common multiple of 15 and 20.

We are looking for the smallest such number, which is the Least Common Multiple (LCM) of 15 and 20.

To find the LCM of 15 and 20, we can use prime factorization.

Prime factorization of 15: $3 \times 5 = 3^1 \times 5^1$

Prime factorization of 20: $2 \times 10 = 2 \times 2 \times 5 = 2^2 \times 5^1$

The LCM is the product of the highest powers of all prime factors that appear in either factorization.

The prime factors involved are 2, 3, and 5.

Highest power of 2: $2^2$ (from 20)

Highest power of 3: $3^1$ (from 15)

Highest power of 5: $5^1$ (from both 15 and 20)

LCM$(15, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 12 \times 5 = 60$.

The smallest number that is divisible by both 15 and 20 is 60.

Let's check the options:

(A) 5: Not divisible by 15 or 20 (it's a common factor).

(B) 30: Divisible by 15 ($15 \times 2 = 30$), but not by 20.

(C) 60: Divisible by 15 ($15 \times 4 = 60$) and divisible by 20 ($20 \times 3 = 60$). This is a common multiple.

(D) 120: Divisible by 15 ($15 \times 8 = 120$) and divisible by 20 ($20 \times 6 = 120$). This is a common multiple, but not the smallest.

The smallest common multiple is 60.

The correct option is (C) 60.

Euclid's Division Lemma:

Given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying $a = bq + r$, where $0 \leq r < b$.

Using Euclid's Division Algorithm to find HCF:

Euclid's Division Algorithm is a technique to compute the Highest Common Factor (HCF) of two positive integers. It is based on Euclid's Division Lemma. The algorithm proceeds as follows:

Step 1: Apply Euclid's Division Lemma to $a$ and $b$ ($a > b$). So, we find whole numbers $q$ and $r$ such that $a = bq + r$, $0 \leq r < b$.

Step 2: If $r = 0$, then $b$ is the HCF of $a$ and $b$.

Step 3: If $r \neq 0$, then apply the lemma to $b$ and $r$.

Step 4: Continue this process until the remainder is zero. The divisor at this stage will be the required HCF.

Finding the HCF of 135 and 225:

We need to find the HCF of 135 and 225 using Euclid's Division Algorithm.

Here, $a = 225$ and $b = 135$.

Step 1: Apply the division lemma to 225 and 135.

The remainder is $90 \neq 0$.

Step 2: Apply the division lemma to the new divisor 135 and the new remainder 90.

The remainder is $45 \neq 0$.

Step 3: Apply the division lemma to the new divisor 90 and the new remainder 45.

The remainder is 0.

Since the remainder is zero, the process stops. The divisor at this stage is 45.

Therefore, the HCF of 135 and 225 is 45.

We need to find the HCF of 867 and 255 using Euclid's Division Algorithm.

Apply Euclid's Division Lemma to 867 and 255. Here, $a = 867$ and $b = 255$.

Step 1: Divide 867 by 255.

The remainder is $102 \neq 0$.

Step 2: Apply the division lemma to the new divisor 255 and the new remainder 102.

The remainder is $51 \neq 0$.

Step 3: Apply the division lemma to the new divisor 102 and the new remainder 51.

The remainder is 0.

Since the remainder is zero, the process stops. The divisor at this stage is 51.

Therefore, the HCF of 867 and 255 is 51.

Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Expressing 3825 as a product of its prime factors:

We will find the prime factors of 3825 using the division method.

Divide 3825 by the smallest prime number that divides it. 3825 ends with 5, so it is divisible by 5. It is also divisible by 3 (sum of digits $3+8+2+5 = 18$, which is divisible by 3).

Let's start with 3:

The prime factors of 3825 are 3, 3, 5, 5, and 17.

Therefore, the prime factorisation of 3825 is:

$3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$

First, we find the prime factorisation of 12 and 18.

Prime factorisation of 12:

$12 = 2 \times 6 = 2 \times 2 \times 3 = 2^2 \times 3^1$

Prime factorisation of 18:

$18 = 2 \times 9 = 2 \times 3 \times 3 = 2^1 \times 3^2$

To find HCF:

The HCF is the product of the smallest power of each common prime factor involved in the numbers.

The common prime factors are 2 and 3.

Smallest power of 2 is $2^1$.

Smallest power of 3 is $3^1$.

HCF$(12, 18) = 2^1 \times 3^1 = 2 \times 3 = 6$

To find LCM:

The LCM is the product of the greatest power of each prime factor involved in the numbers.

The prime factors involved are 2 and 3.

Greatest power of 2 is $2^2$.

Greatest power of 3 is $3^2$.

LCM$(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36$

Thus, the HCF of 12 and 18 is 6 and the LCM of 12 and 18 is 36.

Given:

Two numbers are 18 and 20.

HCF$(18, 20) = 2$

To Find:

LCM$(18, 20)$

Solution:

We know that for any two positive integers $a$ and $b$, the product of their HCF and LCM is equal to the product of the numbers.

That is, HCF$(a, b) \times$ LCM$(a, b) = a \times b$.

Here, $a = 18$ and $b = 20$.

Substitute the given values into the formula:

We are given that HCF$(18, 20) = 2$.

Now, divide both sides by 2 to find the LCM.

Thus, the LCM of 18 and 20 is 180.

We are asked to explain why the number $5 \times 7 \times 11 + 7$ is a composite number.

First, let's evaluate the expression:

So, the number is 392.

Now, let's look at the original expression $5 \times 7 \times 11 + 7$. We can take 7 as a common factor from both terms.

We know that $56 = 7 \times 8 = 7 \times 2^3$.

So, $5 \times 7 \times 11 + 7 = 7 \times 56 = 7 \times 7 \times 8 = 7^2 \times 2^3$.

A composite number is a positive integer that has at least one divisor other than 1 and itself. In other words, a composite number can be formed by multiplying two smaller positive integers.

The number $5 \times 7 \times 11 + 7$ can be expressed as $7 \times 56$.

Since it can be factored into $7$ and $56$, and both 7 and 56 are positive integers other than 1, the number $5 \times 7 \times 11 + 7$ has factors other than 1 and itself (e.g., 7 and 56). Its prime factorisation is $2^3 \times 7^2$.

Therefore, $5 \times 7 \times 11 + 7$ is a composite number.

A rational number $\frac{p}{q}$, where $p$ and $q$ are coprime integers and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ is of the form $2^n 5^m$, where $n$ and $m$ are non-negative integers.

The given rational number is $\frac{13}{125}$.

Here, the numerator is $p = 13$ and the denominator is $q = 125$.

The numbers 13 and 125 are coprime.

Now, we find the prime factorisation of the denominator, $q = 125$.

We can write the prime factorisation of the denominator in the form $2^n 5^m$ as:

Here, $n=0$ and $m=3$, which are non-negative integers.

Since the prime factorisation of the denominator 125 is of the form $2^n 5^m$, the rational number $\frac{13}{125}$ has a terminating decimal expansion.

A rational number $\frac{p}{q}$, where $p$ and $q$ are coprime integers and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ is of the form $2^n 5^m$, where $n$ and $m$ are non-negative integers.

The given rational number is $\frac{29}{343}$.

Here, the numerator is $p = 29$ and the denominator is $q = 343$.

The numbers 29 and 343 are coprime because 29 is a prime number and 343 is $7^3$.

Now, we find the prime factorisation of the denominator, $q = 343$.

The prime factorisation of the denominator 343 is $7^3$.

Since the prime factorisation of the denominator 343 contains the prime factor 7, which is not 2 or 5, the rational number $\frac{29}{343}$ does not have a terminating decimal expansion.

Therefore, the rational number $\frac{29}{343}$ has a non-terminating repeating decimal expansion.

The primary difference between rational and irrational numbers lies in their definition based on fractions and their decimal representations.

Rational Numbers:

A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

In decimal form, rational numbers are either terminating or non-terminating and repeating (recurrent).

Examples of rational numbers:

1. $\frac{1}{2} = 0.5$ (Terminating decimal)

2. $0.\overline{3} = 0.333... = \frac{1}{3}$ (Non-terminating repeating decimal)

Other examples include integers (like $5 = \frac{5}{1}$), terminating decimals (like $1.25 = \frac{125}{100} = \frac{5}{4}$), and fractions.

Irrational Numbers:

An irrational number is a number that cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

In decimal form, irrational numbers are non-terminating and non-repeating.

Examples of irrational numbers:

1. $\sqrt{2} \approx 1.41421356...$ (Non-terminating and non-repeating decimal)

2. $\pi \approx 3.14159265...$ (Non-terminating and non-repeating decimal)

Other examples include $\sqrt{3}$, $\sqrt{5}$, $e$, etc.

We will prove that $\sqrt{2}$ is an irrational number using the method of proof by contradiction.

Assumption:

Let us assume, for the sake of contradiction, that $\sqrt{2}$ is a rational number.

Proof:

If $\sqrt{2}$ is a rational number, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ have no common factors other than 1 (i.e., the fraction $\frac{p}{q}$ is in its simplest form or $p$ and $q$ are coprime).

Square both sides of the equation:

Rearrange the equation:

Equation (i) shows that $p^2$ is an even number because it is a multiple of 2.

We know that if the square of an integer is even, then the integer itself must be even. Therefore, $p$ is an even number.

Since $p$ is even, we can write $p$ as $2k$ for some integer $k$.

Substitute the value of $p$ from equation (ii) into equation (i):

Divide both sides by 2:

Equation (iii) shows that $q^2$ is an even number because it is a multiple of 2.

Again, if the square of an integer is even, then the integer itself must be even. Therefore, $q$ is an even number.

So, we have concluded that both $p$ and $q$ are even numbers.

Contradiction:

If both $p$ and $q$ are even, it means that both $p$ and $q$ have a common factor of 2.

This contradicts our initial assumption that $p$ and $q$ have no common factors other than 1 (i.e., the fraction $\frac{p}{q}$ was in its simplest form).

Conclusion:

Since our assumption that $\sqrt{2}$ is rational leads to a contradiction, the assumption must be false.

Therefore, $\sqrt{2}$ is an irrational number.

We want to show that $3\sqrt{2}$ is irrational, given that $\sqrt{2}$ is irrational.

We will use the method of proof by contradiction.

Assumption:

Let us assume, for the sake of contradiction, that $3\sqrt{2}$ is a rational number.

Proof:

If $3\sqrt{2}$ is a rational number, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Now, we can isolate $\sqrt{2}$ from this equation by dividing both sides by 3:

Since $p$ and $q$ are integers and $q \neq 0$, it follows that $3q$ is also a non-zero integer.

Therefore, the expression $\frac{p}{3q}$ is in the form of an integer divided by a non-zero integer.

By the definition of a rational number, this means that $\frac{p}{3q}$ is a rational number.

So, if $3\sqrt{2}$ is rational, then $\sqrt{2}$ must be rational.

Contradiction:

This conclusion, that $\sqrt{2}$ is rational, contradicts the given information that $\sqrt{2}$ is an irrational number.

Conclusion:

Since our assumption that $3\sqrt{2}$ is rational leads to a contradiction with the given fact that $\sqrt{2}$ is irrational, the assumption must be false.

Therefore, $3\sqrt{2}$ is an irrational number.

To determine if a rational number has a terminating decimal expansion without actual division, we need to check the prime factorisation of its denominator after simplifying the fraction to its lowest terms.

The given rational number is $\frac{6}{15}$.

First, we simplify the fraction by cancelling the common factor, which is 3:

Now, the fraction is in its simplest form, $\frac{p}{q} = \frac{2}{5}$.

The denominator is $q = 5$.

Find the prime factorisation of the denominator:

A rational number $\frac{p}{q}$ (in simplest form) has a terminating decimal expansion if the prime factorisation of the denominator $q$ is of the form $2^n 5^m$, where $n$ and $m$ are non-negative integers.

The prime factorisation of the denominator 5 can be written as $2^0 \times 5^1$.

Here, $n=0$ and $m=1$, which are non-negative integers.

Since the prime factors of the denominator are only 5 (or includes 2 and 5), the decimal expansion is terminating.

To write the decimal expansion of $\frac{6}{15}$ (which is equal to $\frac{2}{5}$):

We can convert the fraction to a decimal by performing division or by making the denominator a power of 10.

The decimal expansion is 0.4.

Conclusion:

The decimal expansion of the rational number $\frac{6}{15}$ is terminating, and its decimal expansion is 0.4.

No, the HCF of two numbers cannot be 14 and their LCM be 204.

Reason:

For any two positive integers, the HCF (Highest Common Factor) must always be a factor of their LCM (Least Common Multiple).

In other words, LCM must be divisible by HCF.

In this case, the given HCF is 14 and the given LCM is 204.

We need to check if 14 is a factor of 204, or if 204 is divisible by 14.

Let's perform the division $204 \div 14$:

When 204 is divided by 14, the remainder is 8, which is not 0.

This means that 204 is not divisible by 14.

Since the HCF (14) is not a factor of the LCM (204), it is not possible for two numbers to have 14 as their HCF and 204 as their LCM.

Let the largest number be $N$.

According to the question, when 615 is divided by $N$, the remainder is 6.

This means that $615 - 6 = 609$ must be perfectly divisible by $N$.

Similarly, when 963 is divided by $N$, the remainder is 6.

This means that $963 - 6 = 957$ must be perfectly divisible by $N$.

We are looking for the largest number $N$ that divides both 609 and 957 exactly.

This largest number is the HCF (Highest Common Factor) of 609 and 957.

We can find the HCF using Euclid's Division Algorithm.

Here, $a = 957$ and $b = 609$.

Step 1: Apply the division lemma to 957 and 609.

The remainder is $348 \neq 0$.

Step 2: Apply the division lemma to the new divisor 609 and the new remainder 348.

The remainder is $261 \neq 0$.

Step 3: Apply the division lemma to the new divisor 348 and the new remainder 261.

The remainder is $87 \neq 0$.

Step 4: Apply the division lemma to the new divisor 261 and the new remainder 87.

The remainder is 0.

Since the remainder is zero, the process stops. The divisor at this stage is 87.

Therefore, the HCF of 609 and 957 is 87.

The largest number that divides 615 and 963, leaving a remainder of 6 in each case, is the HCF of $(615 - 6)$ and $(963 - 6)$, which is HCF$(609, 957)$.

The HCF$(609, 957) = 87$.

We also note that the remainder 6 is less than the divisor 87 ($6 < 87$).

Thus, the required largest number is 87.

We need to find the maximum capacity of a container that can measure the petrol of both tankers exactly. This means the capacity of the container must be a common divisor of the quantities of petrol in both tankers (850 litres and 680 litres).

To find the maximum capacity, we need to find the Highest Common Factor (HCF) of 850 and 680.

We will use Euclid's Division Algorithm to find the HCF.

Apply Euclid's Division Lemma to 850 and 680. Here, $a = 850$ and $b = 680$.

Step 1: Divide 850 by 680.

The remainder is $170 \neq 0$.

Step 2: Apply the division lemma to the new divisor 680 and the new remainder 170.

The remainder is 0.

Since the remainder is zero, the process stops. The divisor at this stage is 170.

Therefore, the HCF of 850 and 680 is 170.

The maximum capacity of the container that can measure the petrol of both tankers when used an exact number of times is the HCF of their capacities.

Maximum capacity = HCF$(850, 680) = 170$ litres.

Thus, the maximum capacity of the container is 170 litres.

For any positive integer to end with the digit 0, it must be divisible by 10.

A number is divisible by 10 if and only if its prime factorisation contains both 2 and 5 as factors.

Consider the number $4^n$, where $n$ is a natural number.

We find the prime factorisation of the base, 4:

Now, let's find the prime factorisation of $4^n$:

The prime factorisation of $4^n$ is $2^{2n}$. This means that the only prime factor in the expansion of $4^n$ is 2.

For $4^n$ to end with the digit 0, its prime factorisation must contain both 2 and 5.

However, the prime factorisation of $4^n$ is $2^{2n}$, which does not include the prime factor 5.

According to the Fundamental Theorem of Arithmetic, the prime factorisation of any composite number is unique.

Since 5 is not a prime factor of $4^n$, $4^n$ is not divisible by 5.

Since $4^n$ is not divisible by 5, it cannot be divisible by 10 (which requires divisibility by both 2 and 5).

Therefore, $4^n$ cannot end with the digit 0 for any natural number $n$.

To find the decimal expansion of a rational number $\frac{p}{q}$ without long division, we can express the denominator $q$ as a power of 10. This is possible if the prime factors of $q$ are only 2 and 5.

The given rational number is $\frac{17}{80}$.

Here, the numerator is $p = 17$ and the denominator is $q = 80$.

First, find the prime factorisation of the denominator 80.

The prime factors of the denominator are 2 and 5. This confirms that the decimal expansion will be terminating.

To convert the fraction to a decimal, we need to make the denominator a power of 10. A power of 10 has the form $2^k \times 5^k$. In the prime factorisation of 80, the power of 2 is 4 and the power of 5 is 1.

To make the powers equal to 4, we need to multiply the denominator by $5^{4-1} = 5^3 = 125$.

We must multiply both the numerator and the denominator by the same number to keep the value of the fraction unchanged.

Now, calculate the numerator and the denominator:

Numerator: $17 \times 125 = 2125$

Denominator: $2^4 \times 5^4 = (2 \times 5)^4 = 10^4 = 10000$

So the fraction becomes:

To write the decimal expansion of $\frac{2125}{10000}$, we move the decimal point 4 places to the left from the right end of the numerator (since there are 4 zeros in the denominator).

Therefore, the decimal expansion of $\frac{17}{80}$ is 0.2125, which is a terminating decimal.

We will find the LCM and HCF of 510 and 92 by first finding their prime factorisations.

Prime factorisation of 510:

We divide 510 by prime numbers:

So, the prime factorisation of 510 is:

Prime factorisation of 92:

We divide 92 by prime numbers:

So, the prime factorisation of 92 is:

To find HCF:

The HCF is the product of the smallest power of each common prime factor in the prime factorisations of the numbers.

Prime factors of 510 are $\{2, 3, 5, 17\}$.

Prime factors of 92 are $\{2, 23\}$.

The only common prime factor is 2.

The smallest power of the common prime factor 2 is $2^1$ (from the factorisation of 510).

To find LCM:

The LCM is the product of the greatest power of each prime factor involved in the prime factorisations of the numbers.

The prime factors involved are 2, 3, 5, 17, and 23.

Greatest power of 2 is $2^2$ (from 92).

Greatest power of 3 is $3^1$ (from 510).

Greatest power of 5 is $5^1$ (from 510).

Greatest power of 17 is $17^1$ (from 510).

Greatest power of 23 is $23^1$ (from 92).

Thus, the HCF of 510 and 92 is 2, and the LCM of 510 and 92 is 23460.

We will use Euclid's Division Lemma to prove this statement.

Euclid's Division Lemma:

Given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying $a = bq + r$, where $0 \leq r < b$.

Proof:

Let $a$ be any positive odd integer.

We apply Euclid's Division Lemma with $a$ as the dividend and $b = 6$ as the divisor.

According to the lemma, there exist unique integers $q$ and $r$ such that:

where $q$ is the quotient and $r$ is the remainder.

The possible values for the remainder $r$ are integers such that $0 \leq r < b$. Since $b = 6$, the possible values for $r$ are $0, 1, 2, 3, 4,$ and $5$.

Thus, any positive integer $a$ can be written in one of the following forms:

• $a = 6q + 0 = 6q$
• $a = 6q + 1$
• $a = 6q + 2$
• $a = 6q + 3$
• $a = 6q + 4$
• $a = 6q + 5$

Now, let's examine which of these forms represent odd integers.

An integer is odd if it is not divisible by 2.

Consider the forms:

• $6q = 2(3q)$. This is a multiple of 2, so it is an even integer.
• $6q + 1 = 2(3q) + 1$. This is of the form $2k + 1$ (where $k=3q$), so it is an odd integer.
• $6q + 2 = 2(3q + 1)$. This is a multiple of 2, so it is an even integer.
• $6q + 3 = 2(3q) + 3 = 2(3q) + 2 + 1 = 2(3q + 1) + 1$. This is of the form $2k + 1$ (where $k=3q+1$), so it is an odd integer.
• $6q + 4 = 2(3q + 2)$. This is a multiple of 2, so it is an even integer.
• $6q + 5 = 2(3q) + 5 = 2(3q) + 4 + 1 = 2(3q + 2) + 1$. This is of the form $2k + 1$ (where $k=3q+2$), so it is an odd integer.

Since $a$ is a positive odd integer, it cannot be of the form $6q$, $6q + 2$, or $6q + 4$ (as these represent even integers).

Therefore, any positive odd integer must be of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.

To find the HCF of three numbers (72, 108, and 180) using Euclid's Division Algorithm, we first find the HCF of any two of the numbers, and then find the HCF of the result and the third number.

Let's first find the HCF of 72 and 108 using Euclid's Division Algorithm.

Apply Euclid's Division Lemma to 108 and 72. Here, $a = 108$ and $b = 72$.

Step 1: Divide 108 by 72.

The remainder is $36 \neq 0$.

Step 2: Apply the division lemma to the new divisor 72 and the new remainder 36.

The remainder is 0.

The last non-zero divisor is 36.

So, HCF$(72, 108) = 36$.

Now, we find the HCF of the result (36) and the third number (180) using Euclid's Division Algorithm.

Apply Euclid's Division Lemma to 180 and 36. Here, $a = 180$ and $b = 36$.

Step 1: Divide 180 by 36.

The remainder is 0.

The last non-zero divisor is 36.

So, HCF$(36, 180) = 36$.

The HCF of 72, 108, and 180 is the HCF of HCF$(72, 108)$ and 180, which is HCF$(36, 180)$.

Therefore, the HCF of 72, 108, and 180 is 36.

First, we need to find the HCF of 65 and 117 using Euclid's Division Algorithm.

Apply Euclid's Division Lemma to 117 and 65. Here, $a = 117$ and $b = 65$.

Step 1: Divide 117 by 65.

The remainder is $52 \neq 0$.

Step 2: Apply the division lemma to the new divisor 65 and the new remainder 52.

The remainder is $13 \neq 0$.

Step 3: Apply the division lemma to the new divisor 52 and the new remainder 13.

The remainder is 0.

The last non-zero divisor is 13. So, the HCF of 65 and 117 is 13.

We are given that the HCF of 65 and 117 is expressible in the form $65m - 117$.

Therefore, we can set the calculated HCF equal to the given form:

Now, we solve this linear equation for $m$.

Add 117 to both sides of the equation:

Divide both sides by 65:

Thus, the value of $m$ is 2.

We want to prove that $\frac{1}{\sqrt{2}}$ is irrational, given that $\sqrt{2}$ is irrational.

We will use the method of proof by contradiction.

Assumption:

Let us assume, for the sake of contradiction, that $\frac{1}{\sqrt{2}}$ is a rational number.

Proof:

If $\frac{1}{\sqrt{2}}$ is a rational number, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Since $q \neq 0$ and $p$ is an integer, if $p=0$, then $\frac{1}{\sqrt{2}} = 0$, which is not possible as $\sqrt{2}$ is a non-zero number. So, $p \neq 0$.

Take the reciprocal of both sides of the equation:

Since $q$ and $p$ are integers and $p \neq 0$, the expression $\frac{q}{p}$ is in the form of an integer divided by a non-zero integer.

By the definition of a rational number, this means that $\frac{q}{p}$ is a rational number.

Therefore, $\sqrt{2}$ must be a rational number.

Contradiction:

This conclusion, that $\sqrt{2}$ is rational, contradicts the given information in the question that $\sqrt{2}$ is an irrational number.

Conclusion:

Since our assumption that $\frac{1}{\sqrt{2}}$ is rational leads to a contradiction with the given fact that $\sqrt{2}$ is irrational, the assumption must be false.

Therefore, $\frac{1}{\sqrt{2}}$ is an irrational number.

For a natural number to end with the digit 0, it must be divisible by 10.

A number is divisible by 10 if and only if its prime factorisation contains both 2 and 5 as factors.

Consider the number $12^n$, where $n$ is a natural number.

We find the prime factorisation of the base, 12:

Now, let's find the prime factorisation of $12^n$ by raising the prime factors of 12 to the power of $n$:

The prime factorisation of $12^n$ is $2^{2n} \times 3^n$. This shows that the only prime factors in the expansion of $12^n$ are 2 and 3.

For $12^n$ to end with the digit 0, its prime factorisation must include the prime factor 5, in addition to 2.

However, the prime factorisation of $12^n$, which is $2^{2n} \times 3^n$, does not contain the prime factor 5.

According to the Fundamental Theorem of Arithmetic, the prime factorisation of any composite number is unique (apart from the order of the factors).

Since 5 is not a prime factor of $12^n$ in its unique prime factorisation, $12^n$ is not divisible by 5.

Since $12^n$ is not divisible by 5, it cannot be divisible by 10 (which requires divisibility by both 2 and 5).

Therefore, $12^n$ cannot end with the digit 0 for any natural number $n$.

Euclid's Division Lemma:

Given two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \leq r < b$. Here, $q$ is the quotient and $r$ is the remainder.

Euclid's Division Algorithm:

This algorithm is a method to find the HCF of two positive integers $a$ and $b$ ($a > b$). It involves applying Euclid's Division Lemma repeatedly until the remainder becomes 0.

The steps are as follows:

Step 1: Apply the division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = bq + r$, $0 \leq r < b$.

Step 2: If $r = 0$, then $b$ is the HCF of $a$ and $b$.

Step 3: If $r \neq 0$, replace $a$ by $b$ and $b$ by $r$. Then apply the division lemma to the new pair of numbers.

Step 4: Continue this process until the remainder is zero. The divisor at the stage where the remainder is zero will be the HCF of the original two numbers.

Finding the HCF of 4052 and 12576:

We need to find the HCF of 4052 and 12576 using Euclid's Division Algorithm.

Here, $a = 12576$ and $b = 4052$.

Step 1: Apply the division lemma to 12576 and 4052.

Step 2: Apply the division lemma to the new divisor 4052 and the new remainder 420.

Step 3: Apply the division lemma to the new divisor 420 and the new remainder 272.

Step 4: Apply the division lemma to the new divisor 272 and the new remainder 148.

Step 5: Apply the division lemma to the new divisor 148 and the new remainder 124.

Step 6: Apply the division lemma to the new divisor 124 and the new remainder 24.

Step 7: Apply the division lemma to the new divisor 24 and the new remainder 4.

Since the remainder is zero, the process stops. The divisor at this stage is 4.

Therefore, the HCF of 4052 and 12576 is 4.

The Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

To find the HCF and LCM of $144$ and $192$, we first find their prime factorisation using the Fundamental Theorem of Arithmetic.

Prime factorisation of $144$:

$\begin{array}{c|cc} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$.

Prime factorisation of $192$:

$\begin{array}{c|cc} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$.

Now, we find the HCF and LCM using the prime factorisations.

HCF (Highest Common Factor): The HCF is the product of the smallest powers of each common prime factor.

The common prime factors are $2$ and $3$.

Smallest power of $2$ is $2^4$.

Smallest power of $3$ is $3^1$.

HCF$(144, 192) = 2^4 \times 3^1 = 16 \times 3 = 48$.

LCM (Lowest Common Multiple): The LCM is the product of the highest powers of all prime factors involved in the numbers.

The prime factors involved are $2$ and $3$.

Highest power of $2$ is $2^6$.

Highest power of $3$ is $3^2$.

LCM$(144, 192) = 2^6 \times 3^2 = 64 \times 9 = 576$.

Now, we verify that HCF $\times$ LCM = product of the two numbers.

Product of the two numbers $= 144 \times 192$.

$144 \times 192 = 27648$.

Product of HCF and LCM $= 48 \times 576$.

$48 \times 576 = 27648$.

Since, $144 \times 192 = 27648$ and HCF $\times$ LCM $= 27648$, the verification is complete.

HCF $\times$ LCM = Product of the two numbers.

Proof:

To prove that $\sqrt{5}$ is an irrational number, we will use the method of contradiction.

Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.

According to the definition of a rational number, if $\sqrt{5}$ is rational, then it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime (meaning they have no common factors other than $1$).

So, we assume:

where $a, b \in \mathbb{Z}$, $b \neq 0$, and HCF$(a, b) = 1$.

Squaring both sides of equation (i), we get:

$(\sqrt{5})^2 = \left(\frac{a}{b}\right)^2$

$5 = \frac{a^2}{b^2}$

Rearranging the equation, we get:

Equation (ii) implies that $a^2$ is a multiple of $5$.

According to the theorem that if a prime number $p$ divides $a^2$, then $p$ divides $a$, since $5$ is a prime number and $5$ divides $a^2$, it must be that $5$ divides $a$.

So, we can write $a = 5c$ for some integer $c$.

Now, substitute $a = 5c$ into equation (ii):

$(5c)^2 = 5b^2$

$25c^2 = 5b^2$

Divide both sides by $5$:

Equation (iii) implies that $b^2$ is a multiple of $5$.

Again, using the theorem that if a prime number $p$ divides $b^2$, then $p$ divides $b$, since $5$ is a prime number and $5$ divides $b^2$, it must be that $5$ divides $b$.

So, we have concluded that $5$ divides $a$ and $5$ divides $b$. This means that $5$ is a common factor of both $a$ and $b$.

However, this contradicts our initial assumption that $a$ and $b$ are coprime (i.e., they have no common factors other than $1$).

Since our assumption that $\sqrt{5}$ is rational leads to a contradiction, the assumption must be false.

Therefore, $\sqrt{5}$ is an irrational number.

A rational number is defined as a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

When we convert a rational number into its decimal form, the decimal expansion will be either terminating or non-terminating repeating.

A decimal expansion is terminating if the digits after the decimal point end after a finite number of steps (e.g., $0.25$, $3.785$).

A decimal expansion is non-terminating repeating (or recurring) if the digits after the decimal point continue infinitely, but a block of digits repeats itself regularly (e.g., $0.333... = 0.\overline{3}$, $1.272727... = 1.\overline{27}$).

Condition for a rational number to have a terminating decimal expansion:

A rational number $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.

Explanation of the condition:

A terminating decimal can always be written as a fraction with a denominator that is a power of $10$ (e.g., $0.25 = \frac{25}{100}$, $3.785 = \frac{3785}{1000}$).

The prime factorization of $10$ is $2 \times 5$. Therefore, any positive integer power of $10$ will have a prime factorization consisting only of powers of $2$ and $5$ ($10^k = (2 \times 5)^k = 2^k \times 5^k$).

If the denominator $q$ of a rational number $\frac{p}{q}$ has a prime factorization of the form $2^m \times 5^n$, we can multiply the numerator and denominator by appropriate powers of $2$ and $5$ to make the denominator a power of $10$.

If $m > n$, multiply by $5^{m-n}$: $\frac{p}{2^m 5^n} = \frac{p \times 5^{m-n}}{(2^m 5^n) \times 5^{m-n}} = \frac{p \times 5^{m-n}}{2^m 5^m} = \frac{p \times 5^{m-n}}{10^m}$.

If $n > m$, multiply by $2^{n-m}$: $\frac{p}{2^m 5^n} = \frac{p \times 2^{n-m}}{(2^m 5^n) \times 2^{n-m}} = \frac{p \times 2^{n-m}}{2^n 5^n} = \frac{p \times 2^{n-m}}{10^n}$.

If $m=n$, the denominator is already a power of $10$: $\frac{p}{2^m 5^m} = \frac{p}{10^m}$.

In all these cases, the fraction is converted to an equivalent fraction with a denominator that is a power of $10$, which results in a terminating decimal expansion.

If the denominator $q$ contains any prime factor other than $2$ or $5$, it is impossible to convert the fraction into an equivalent fraction with a denominator that is a power of $10$. In such cases, the decimal expansion will be non-terminating and repeating.

Decimal expansion of $\frac{13}{3125}$:

First, find the prime factorization of the denominator $3125$.

$\begin{array}{c|cc} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.

The denominator is of the form $2^0 \times 5^5$. Since the prime factors are only $5$ (and effectively $2^0$), the decimal expansion will be terminating.

To convert it to a decimal, we can make the denominator a power of $10$. We need to multiply $5^5$ by $2^5$.

$\frac{13}{3125} = \frac{13}{5^5} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{(5 \times 2)^5} = \frac{416}{10^5} = \frac{416}{100000}$.

Now, we can easily write the decimal expansion:

$\frac{416}{100000} = 0.00416$.

Decimal expansion of $\frac{17}{8}$:

First, find the prime factorization of the denominator $8$.

$\begin{array}{c|cc} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $8 = 2 \times 2 \times 2 = 2^3$.

The denominator is of the form $2^3 \times 5^0$. Since the prime factors are only $2$ (and effectively $5^0$), the decimal expansion will be terminating.

To convert it to a decimal, we can make the denominator a power of $10$. We need to multiply $2^3$ by $5^3$.

$\frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{2^3 \times 5^3} = \frac{17 \times 125}{(2 \times 5)^3} = \frac{2125}{10^3} = \frac{2125}{1000}$.

Now, we can easily write the decimal expansion:

$\frac{2125}{1000} = 2.125$.

Proof:

We are given that $\sqrt{5}$ is an irrational number.

We need to prove that $3 + 2\sqrt{5}$ is an irrational number.

Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is a rational number.

According to the definition of a rational number, if $3 + 2\sqrt{5}$ is rational, then it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

So, we assume:

where $a, b \in \mathbb{Z}$ and $b \neq 0$.

Now, we manipulate equation (i) to isolate $\sqrt{5}$ on one side.

Subtract $3$ from both sides:

$2\sqrt{5} = \frac{a}{b} - 3$

$2\sqrt{5} = \frac{a - 3b}{b}$

Divide both sides by $2$:

Now, let's analyze the right-hand side of equation (ii).

Since $a$ and $b$ are integers, $a - 3b$ is also an integer.

Since $b$ is a non-zero integer, $2b$ is also a non-zero integer.

Therefore, the expression $\frac{a - 3b}{2b}$ is in the form of an integer divided by a non-zero integer, which is the definition of a rational number.

So, equation (ii) implies that $\sqrt{5}$ is a rational number.

However, we are given that $\sqrt{5}$ is an irrational number.

This creates a contradiction: $\sqrt{5}$ cannot be both rational and irrational.

The contradiction arose because of our initial assumption that $3 + 2\sqrt{5}$ is a rational number.

Therefore, our assumption must be false.

Hence, $3 + 2\sqrt{5}$ is an irrational number.

To Find: The time after which Sonia and Ravi will meet again at the starting point.

Sonia takes 18 minutes to complete one round.

Ravi takes 12 minutes to complete one round.

They start at the same point, at the same time, and go in the same direction.

They will meet again at the starting point when the time elapsed is a multiple of both 18 minutes (Sonia's time per round) and 12 minutes (Ravi's time per round).

The earliest time they will meet again at the starting point is the least common multiple (LCM) of 18 and 12.

We find the LCM of 18 and 12 using prime factorization.

Prime factorization of 18:

$\begin{array}{c|cc} 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$18 = 2 \times 3 \times 3 = 2^1 \times 3^2$

Prime factorization of 12:

$\begin{array}{c|cc} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$12 = 2 \times 2 \times 3 = 2^2 \times 3^1$

To find the LCM, we take the highest power of each prime factor present in the factorizations of 18 and 12.

The prime factors are 2 and 3.

Highest power of 2 is $2^2 = 4$.

Highest power of 3 is $3^2 = 9$.

LCM$(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.

Alternatively, using the LCM calculation format:

$\begin{array}{c|cc} 2 & 18 \;, & 12 \\ \hline 3 & 9 \; , & 6 \\ \hline & 3 \; , & 2 \end{array}$

LCM$(18, 12) = 2 \times 3 \times 3 \times 2 = 36$.

The LCM of 18 and 12 is 36.

This means that after 36 minutes, both Sonia and Ravi will be back at the starting point simultaneously.

Sonia will have completed $36 \div 18 = 2$ rounds.

Ravi will have completed $36 \div 12 = 3$ rounds.

Answer: They will meet again at the starting point after 36 minutes.

Proof:

Let $a$ be any positive integer.

According to Euclid's Division Lemma, for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \leq r < b$.

Let us apply Euclid's Division Lemma to the positive integer $a$ with the divisor $b=3$.

So, $a = 3q + r$, where $q$ is an integer ($q \geq 0$ since $a$ is positive) and $r$ is the remainder such that $0 \leq r < 3$.

The possible values for the remainder $r$ are $0$, $1$, or $2$.

We need to find the square of $a$, i.e., $a^2$, for each possible value of $r$.

Case 1: When $r = 0$

If $r = 0$, then $a = 3q$.

Squaring both sides, we get:

$a^2 = (3q)^2 = 9q^2$

We can rewrite this as:

$a^2 = 3(3q^2)$

Let $m = 3q^2$. Since $q$ is an integer, $q^2$ is an integer, and therefore $3q^2$ is also an integer. So, $m$ is an integer.

Thus, $a^2$ is of the form $3m$ for some integer $m$.

Case 2: When $r = 1$

If $r = 1$, then $a = 3q + 1$.

Squaring both sides, we get:

$a^2 = (3q + 1)^2$

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$a^2 = (3q)^2 + 2(3q)(1) + 1^2$

$a^2 = 9q^2 + 6q + 1$

We can factor out $3$ from the first two terms:

$a^2 = 3(3q^2 + 2q) + 1$

Let $m = 3q^2 + 2q$. Since $q$ is an integer, $3q^2 + 2q$ is also an integer. So, $m$ is an integer.

Thus, $a^2$ is of the form $3m + 1$ for some integer $m$.

Case 3: When $r = 2$

If $r = 2$, then $a = 3q + 2$.

Squaring both sides, we get:

$a^2 = (3q + 2)^2$

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$a^2 = (3q)^2 + 2(3q)(2) + 2^2$

$a^2 = 9q^2 + 12q + 4$

We can rewrite $4$ as $3 + 1$ to factor out $3$:

$a^2 = 9q^2 + 12q + 3 + 1$

$a^2 = 3(3q^2 + 4q + 1) + 1$

Let $m = 3q^2 + 4q + 1$. Since $q$ is an integer, $3q^2 + 4q + 1$ is also an integer. So, $m$ is an integer.

Thus, $a^2$ is of the form $3m + 1$ for some integer $m$.

From all the cases ($r=0$, $r=1$, and $r=2$), we observe that the square of any positive integer $a$ is always of the form $3m$ or $3m + 1$ for some integer $m$.

This concludes the proof.

We need to find the HCF and LCM of the numbers 6, 72, and 120 using the prime factorisation method.

First, we find the prime factorisation of each number:

Prime factorisation of 6:

$\begin{array}{c|cc} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$6 = 2 \times 3 = 2^1 \times 3^1$

Prime factorisation of 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$

Prime factorisation of 120:

$\begin{array}{c|cc} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1$

HCF (Highest Common Factor): The HCF is the product of the smallest powers of each common prime factor among the numbers.

The common prime factors are $2$ and $3$.

Smallest power of $2$ (from $2^1, 2^3, 2^3$) is $2^1$.

Smallest power of $3$ (from $3^1, 3^2, 3^1$) is $3^1$.

The prime factor $5$ is not common to all three numbers.

HCF$(6, 72, 120) = 2^1 \times 3^1 = 2 \times 3 = 6$.

LCM (Lowest Common Multiple): The LCM is the product of the highest powers of all prime factors involved in the numbers.

The prime factors involved are $2$, $3$, and $5$.

Highest power of $2$ (from $2^1, 2^3, 2^3$) is $2^3$.

Highest power of $3$ (from $3^1, 3^2, 3^1$) is $3^2$.

Highest power of $5$ (from $5^0, 5^0, 5^1$) is $5^1$.

LCM$(6, 72, 120) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 72 \times 5 = 360$.

Verification of the relation HCF $\times$ LCM = product of the numbers:

Let's calculate the product of the three numbers:

Product of numbers $= 6 \times 72 \times 120$

$6 \times 72 = 432$

$432 \times 120 = 51840$.

Product of HCF and LCM:

HCF $\times$ LCM $= 6 \times 360$

$6 \times 360 = 2160$.

Comparing the two results:

Product of numbers $= 51840$

HCF $\times$ LCM $= 2160$

Since $51840 \neq 2160$, the relation HCF $\times$ LCM = product of the numbers does not hold true for these three numbers.

Condition for which the property HCF $\times$ LCM = product of the numbers holds:

The property that the product of the HCF and LCM of numbers is equal to the product of the numbers holds true only for two positive integers.

For any two positive integers $a$ and $b$, it is true that:

HCF$(a, b) \times$ LCM$(a, b) = a \times b$.

This property is not generally true for three or more numbers.

To convert a non-terminating repeating decimal expansion into the form $\frac{p}{q}$ (a rational number), we use an algebraic method involving equations and powers of $10$.

Here's the general procedure:

Step 1: Let the given non-terminating repeating decimal be equal to a variable, say $x$.

Step 2: Identify the number of digits in the repeating block (the part under the bar) and the number of digits in the non-repeating part after the decimal point but before the repeating block.

Step 3: Multiply the equation from Step 1 by $10^k$, where $k$ is the number of digits in the non-repeating part. This moves the decimal point just before the repeating block.

Step 4: Multiply the original equation (or the equation from Step 3, depending on the decimal type) by $10^{k+r}$, where $r$ is the number of digits in the repeating block. This moves the decimal point after one full cycle of the repeating block.

Step 5: Subtract the equation where the decimal is just before the repeating block (from Step 3) from the equation where the decimal is after the first repeating block (from Step 4). This subtraction eliminates the repeating part of the decimal.

Step 6: Solve the resulting simple linear equation for $x$. The solution will be in the form of a fraction $\frac{p}{q}$.

Step 7: Simplify the fraction to its lowest terms if necessary.

Let's apply this method to the given examples.

Example 1: Convert $0.1\overline{6}$ to $\frac{p}{q}$.

Let $x = 0.1\overline{6}$. This can be written as $x = 0.16666...$

The non-repeating part after the decimal is '1' (one digit). So, we multiply equation (1) by $10^1 = 10$.

The repeating part is '6' (one digit). So, we multiply equation (1) by $10^{1+1} = 10^2 = 100$.

Now, subtract equation (2) from equation (3) to eliminate the repeating part.

$(3) - (2):$

$100x - 10x = (16.6666...) - (1.6666...)$

$90x = 15$

Solve for $x$:

$x = \frac{15}{90}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 15.

$x = \frac{\cancel{15}^1}{\cancel{90}_6} = \frac{1}{6}$

Thus, $0.1\overline{6} = \frac{1}{6}$.

Example 2: Convert $0.2\overline{35}$ to $\frac{p}{q}$.

Let $x = 0.2\overline{35}$. This can be written as $x = 0.2353535...$

The non-repeating part after the decimal is '2' (one digit). So, we multiply equation (1) by $10^1 = 10$.

The repeating part is '35' (two digits). So, we multiply equation (1) by $10^{1+2} = 10^3 = 1000$.

Now, subtract equation (2) from equation (3) to eliminate the repeating part.

$(3) - (2):$

$1000x - 10x = (235.353535...) - (2.353535...)$

$990x = 233$

Solve for $x$:

$x = \frac{233}{990}$

The numerator 233 is a prime number, and the denominator 990 has prime factors 2, 3, 5, 11. There are no common factors other than 1.

Thus, the fraction is already in its lowest terms.

Therefore, $0.2\overline{35} = \frac{233}{990}$.

Given: $\sqrt{5}$ is an irrational number.

To Prove: $7 - \sqrt{5}$ is an irrational number.

Proof:

To prove that $7 - \sqrt{5}$ is an irrational number, we will use the method of contradiction.

Let us assume, to the contrary, that $7 - \sqrt{5}$ is a rational number.

According to the definition of a rational number, if $7 - \sqrt{5}$ is rational, then it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

So, we assume:

where $a, b \in \mathbb{Z}$ and $b \neq 0$.

Now, we manipulate equation (i) to isolate $\sqrt{5}$ on one side.

Rearrange the equation to move $\sqrt{5}$ to the right side and $\frac{a}{b}$ to the left side:

$7 - \frac{a}{b} = \sqrt{5}$

Combine the terms on the left side:

Now, let's analyze the left-hand side of equation (ii).

Since $a$ and $b$ are integers and $b \neq 0$, $7b$ is an integer. The difference of two integers, $7b - a$, is also an integer.

Since $b$ is a non-zero integer, the denominator $b$ is a non-zero integer.

Therefore, the expression $\frac{7b - a}{b}$ is in the form of an integer divided by a non-zero integer, which is the definition of a rational number.

So, equation (ii) implies that $\sqrt{5}$ is a rational number.

However, we are given that $\sqrt{5}$ is an irrational number.

This creates a contradiction: $\sqrt{5}$ cannot be both rational and irrational.

The contradiction arose because of our initial assumption that $7 - \sqrt{5}$ is a rational number.

Therefore, our assumption must be false.

Hence, $7 - \sqrt{5}$ is an irrational number.

To Find: The maximum number of columns in which the two groups (army contingent and army band) can march.

The number of members in the army contingent is 616.

The number of members in the army band is 32.

They need to march in the same number of columns.

The maximum number of columns will be the largest number that can divide both 616 and 32 exactly. This is the Highest Common Factor (HCF) of 616 and 32.

We will use Euclid's Division Algorithm to find the HCF of 616 and 32.

Euclid's Division Algorithm states that for any two positive integers $a$ and $b$, with $a > b$, we can find integers $q$ and $r$ such that $a = bq + r$, where $0 \leq r < b$. The HCF of $a$ and $b$ is the same as the HCF of $b$ and $r$. We continue this process until the remainder is 0. The divisor at the last step where the remainder is 0 is the HCF.

Let $a = 616$ and $b = 32$.

Step 1: Divide 616 by 32.

$616 = 32 \times q + r$

We perform the division:

$\begin{array}{r} 19 \\ 32{\overline{\smash{\big)}\,616}} \\ \underline{-~32\phantom{0}} \\ 296 \\ \underline{-~288} \\ 8 \end{array}$

So, $616 = 32 \times 19 + 8$.

The remainder is $8$, which is not 0.

Step 2: Now, apply Euclid's algorithm to the divisor from Step 1 (32) and the remainder (8).

Let the new dividend be 32 and the new divisor be 8.

Divide 32 by 8:

$32 = 8 \times q + r$

$\begin{array}{r} 4 \\ 8{\overline{\smash{\big)}\,32}} \\ \underline{-~32} \\ 0 \end{array}$

So, $32 = 8 \times 4 + 0$.

The remainder is $0$. The algorithm stops here.

The divisor at this last step is 8. Therefore, the HCF of 616 and 32 is 8.

HCF$(616, 32) = 8$.

The maximum number of columns in which they can march is the HCF of 616 and 32.

Answer: The maximum number of columns is 8.

This means the army band will march in 8 columns with $32 \div 8 = 4$ members per column, and the army contingent will march in 8 columns with $616 \div 8 = 77$ members per column.

Proof:

Let $a$ be any positive integer.

According to Euclid's Division Lemma, for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \leq r < b$.

Let us apply Euclid's Division Lemma to the positive integer $a$ with the divisor $b=3$.

So, $a = 3q + r$, where $q$ is an integer ($q \geq 0$ since $a$ is positive) and $r$ is the remainder such that $0 \leq r < 3$.

The possible values for the remainder $r$ are $0$, $1$, or $2$.

We need to find the cube of $a$, i.e., $a^3$, for each possible value of $r$.

Case 1: When $r = 0$

If $r = 0$, then $a = 3q$.

Cubing both sides, we get:

$a^3 = (3q)^3 = 27q^3$

We can rewrite this as:

$a^3 = 9(3q^3)$

Let $m = 3q^3$. Since $q$ is an integer, $q^3$ is an integer, and therefore $3q^3$ is also an integer. So, $m$ is an integer.

Thus, $a^3$ is of the form $9m$ for some integer $m$.

Case 2: When $r = 1$

If $r = 1$, then $a = 3q + 1$.

Cubing both sides, we get:

$a^3 = (3q + 1)^3$

Using the identity $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:

$a^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3$

$a^3 = 27q^3 + 3(9q^2) + 9q + 1$

$a^3 = 27q^3 + 27q^2 + 9q + 1$

We can factor out $9$ from the first three terms:

$a^3 = 9(3q^3 + 3q^2 + q) + 1$

Let $m = 3q^3 + 3q^2 + q$. Since $q$ is an integer, $3q^3 + 3q^2 + q$ is also an integer. So, $m$ is an integer.

Thus, $a^3$ is of the form $9m + 1$ for some integer $m$.

Case 3: When $r = 2$

If $r = 2$, then $a = 3q + 2$.

Cubing both sides, we get:

$a^3 = (3q + 2)^3$

Using the identity $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:

$a^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3$

$a^3 = 27q^3 + 3(9q^2)(2) + 3(3q)(4) + 8$

$a^3 = 27q^3 + 54q^2 + 36q + 8$

We can factor out $9$ from the first three terms:

$a^3 = 9(3q^3 + 6q^2 + 4q) + 8$

Let $m = 3q^3 + 6q^2 + 4q$. Since $q$ is an integer, $3q^3 + 6q^2 + 4q$ is also an integer. So, $m$ is an integer.

Thus, $a^3$ is of the form $9m + 8$ for some integer $m$.

From all the cases ($r=0$, $r=1$, and $r=2$), we observe that the cube of any positive integer $a$ is always of the form $9m$, $9m + 1$, or $9m + 8$ for some integer $m$.

This concludes the proof.