Chapter 3 Finding Common Ground (Class 7 - Latest Maths NCERT (Ganita Prakash II) Solutions)

(a) Factors of 90:

To find the factors of 90, we first perform its prime factorisation.

$\begin{array}{c|cc} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorisation of 90 is $2 \times 3 \times 3 \times 5$.

By finding all possible products of these prime factors, we list all the factors of 90:

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

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(b) Factors of 105:

First, we perform the prime factorisation of 105.

$\begin{array}{c|cc} 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factorisation of 105 is $3 \times 5 \times 7$.

By finding all possible products of these prime factors, we list all the factors of 105:

1, 3, 5, 7, 15, 21, 35, 105.

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(c) Factors of 132:

First, we perform the prime factorisation of 132.

$\begin{array}{c|cc} 2 & 132 \\ \hline 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

The prime factorisation of 132 is $2 \times 2 \times 3 \times 11$.

By finding all possible products of these prime factors, we list all the factors of 132:

1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.

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(d) Factors of 360:

First, we perform the prime factorisation of 360.

$\begin{array}{c|cc} 2 & 360 \\ \hline 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorisation of 360 is $2 \times 2 \times 2 \times 3 \times 3 \times 5$.

As per the question, 360 has 24 factors. We list them below:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.

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(e) Factors of 840:

First, we perform the prime factorisation of 840.

$\begin{array}{c|cc} 2 & 840 \\ \hline 2 & 420 \\ \hline 2 & 210 \\ \hline 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factorisation of 840 is $2 \times 2 \times 2 \times 3 \times 5 \times 7$.

As per the question, 840 has 32 factors. We list them below:

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

Given: Pairs of numbers (a) $50, 60$; (b) $140, 275$; (c) $77, 725$; (d) $370, 592$; (e) $81, 243$.

To Find: The common factors and the Highest Common Factor (HCF) for each pair.

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Solution:

To find the HCF and common factors, we will use the Prime Factorisation method for each number.

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(a) Numbers: $50$ and $60$

Prime factorisation of $50$ and $60$:

$\begin{array}{c|cc} 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors of $50 = 2 \times 5 \times 5$

Prime factors of $60 = 2 \times 2 \times 3 \times 5$

The common prime factors are $2$ and $5$. By multiplying these, we get the HCF.

Common factors: $1, 2, 5, 10$

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(b) Numbers: $140$ and $275$

Prime factorisation of $140$ and $275$:

$\begin{array}{c|cc} 2 & 140 \\ \hline 2 & 70 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 5 & 275 \\ \hline 5 & 55 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

Prime factors of $140 = 2 \times 2 \times 5 \times 7$

Prime factors of $275 = 5 \times 5 \times 11$

The only common prime factor is $5$.

Common factors: $1, 5$

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(c) Numbers: $77$ and $725$

Prime factorisation of $77$ and $725$:

$\begin{array}{c|cc} 7 & 77 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 5 & 725 \\ \hline 5 & 145 \\ \hline 29 & 29 \\ \hline & 1 \end{array}$

Prime factors of $77 = 7 \times 11$

Prime factors of $725 = 5 \times 5 \times 29$

There are no common prime factors other than $1$.

Common factors: $1$

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(d) Numbers: $370$ and $592$

Prime factorisation of $370$ and $592$:

$\begin{array}{c|cc} 2 & 370 \\ \hline 5 & 185 \\ \hline 37 & 37 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 2 & 592 \\ \hline 2 & 296 \\ \hline 2 & 148 \\ \hline 2 & 74 \\ \hline 37 & 37 \\ \hline & 1 \end{array}$

Prime factors of $370 = 2 \times 5 \times 37$

Prime factors of $592 = 2 \times 2 \times 2 \times 2 \times 37$

The common prime factors are $2$ and $37$.

Common factors: $1, 2, 37, 74$

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(e) Numbers: $81$ and $243$

Prime factorisation of $81$ and $243$:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Prime factors of $81 = 3 \times 3 \times 3 \times 3$

Prime factors of $243 = 3 \times 3 \times 3 \times 3 \times 3$

The common prime factors are $3 \times 3 \times 3 \times 3$.

Common factors: $1, 3, 9, 27, 81$

(a) $24, 180$

Given: Two numbers $24$ and $180$.

Solution: Let us find the prime factorisation of both numbers:

$\begin{array}{c|cc} 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors of $24 = \mathbf{2} \times \mathbf{2} \times 2 \times \mathbf{3}$

Prime factors of $180 = \mathbf{2} \times \mathbf{2} \times \mathbf{3} \times 3 \times 5$

The common prime factors are $2, 2, \text{ and } 3$.

$\text{HCF} = 2 \times 2 \times 3 = 12$

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(b) $42, 75, 24$

Given: Three numbers $42, 75, \text{ and } 24$.

Solution: Let us find the prime factorisation of the numbers:

$\begin{array}{c|cc} 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$       $\begin{array}{c|cc} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$       $\begin{array}{c|cc} 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Prime factors of $42 = 2 \times \mathbf{3} \times 7$

Prime factors of $75 = \mathbf{3} \times 5 \times 5$

Prime factors of $24 = 2 \times 2 \times 2 \times \mathbf{3}$

The only common prime factor among all three numbers is $3$.

$\text{HCF} = 3$

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(c) $240, 378$

Given: Two numbers $240$ and $378$.

Solution: Let us perform the prime factorisation:

$\begin{array}{c|cc} 2 & 240 \\ \hline 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 2 & 378 \\ \hline 3 & 189 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

Prime factors of $240 = \mathbf{2} \times 2 \times 2 \times 2 \times \mathbf{3} \times 5$

Prime factors of $378 = \mathbf{2} \times \mathbf{3} \times 3 \times 3 \times 7$

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

$\text{HCF} = 2 \times 3 = 6$

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(d) $400, 2500$

Given: Two numbers $400$ and $2500$.

Solution: Let us find the prime factorisation:

$\begin{array}{c|cc} 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors of $400 = \mathbf{2} \times \mathbf{2} \times 2 \times 2 \times \mathbf{5} \times \mathbf{5}$

Prime factors of $2500 = \mathbf{2} \times \mathbf{2} \times \mathbf{5} \times \mathbf{5} \times 5 \times 5$

The common prime factors are $2, 2, 5, \text{ and } 5$.

$\text{HCF} = 2 \times 2 \times 5 \times 5 = 100$

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(e) $300, 800$

Given: Two numbers $300$ and $800$.

Solution: Performing prime factorisation:

$\begin{array}{c|cc} 2 & 300 \\ \hline 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$            $\begin{array}{c|cc} 2 & 800 \\ \hline 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors of $300 = \mathbf{2} \times \mathbf{2} \times 3 \times \mathbf{5} \times \mathbf{5}$

Prime factors of $800 = \mathbf{2} \times \mathbf{2} \times 2 \times 2 \times 2 \times \mathbf{5} \times \mathbf{5}$

The common prime factors are $2, 2, 5, \text{ and } 5$.

$\text{HCF} = 2 \times 2 \times 5 \times 5 = 100$

Given:

Two numbers $72$ and $144$ are factorised as:

$72 = 6 \times 12$

$144 = 8 \times 18$

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To Find:

Whether it can be said that these two numbers have no common factor other than $1$ and the reason why.

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Solution:

No, one cannot say that these two numbers have no common factor other than $1$.

The reason is that the factors provided ($6, 12, 8,$ and $18$) are composite numbers. Composite numbers can be further broken down into their prime factors. Even though the specific factors in the given product representations ($6 \times 12$ and $8 \times 18$) appear different, they may share the same prime building blocks.

Let us look at the Prime Factorisation of both numbers:

For 72:

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

So, $72 = 2 \times 2 \times 2 \times 3 \times 3$

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For 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, $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

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Conclusion:

By comparing the prime factorisations, we can see that they share several common prime factors ($2$ and $3$). In fact, 72 itself is a factor of 144 ($144 = 72 \times 2$). Therefore, they have many common factors such as $2, 3, 4, 6, 8, 9, 12, 18, 24, 36$ and $72$.

Simply looking at one specific pair of composite factors does not give the complete picture of all the factors a number possesses.

(a) LCM of $30, 72$

To find the Least Common Multiple (LCM), we use the common division method:

$\begin{array}{c|cc} 2 & 30 \;, & 72 \\ \hline 2 & 15 \;, & 36 \\ \hline 2 & 15 \;, & 18 \\ \hline 3 & 15 \;, & 9 \\ \hline & 5 \;, & 3 \end{array}$

The LCM is the product of all the divisors and the remaining numbers in the last row.

$\text{LCM} = 2 \times 2 \times 2 \times 3 \times 5 \times 3$

$\text{LCM} = 8 \times 15 \times 3$

$\text{LCM} = \mathbf{360}$

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(b) LCM of $36, 54$

Using the common division method:

$\begin{array}{c|cc} 2 & 36 \;, & 54 \\ \hline 3 & 18 \;, & 27 \\ \hline 3 & 6 \;, & 9 \\ \hline & 2 \;, & 3 \end{array}$

$\text{LCM} = 2 \times 3 \times 3 \times 2 \times 3$

$\text{LCM} = 18 \times 6$

$\text{LCM} = \mathbf{108}$

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(c) LCM of $105, 195, 65$

Using the common division method for three numbers:

$\begin{array}{c|cc} 5 & 105 \;, & 195 \;, & 65 \\ \hline 3 & 21 \;, & 39 \;, & 13 \\ \hline 13 & 7 \;, & 13 \;, & 13 \\ \hline & 7 \;, & 1 \;, & 1 \end{array}$

$\text{LCM} = 5 \times 3 \times 13 \times 7 \times 1 \times 1$

$\text{LCM} = 15 \times 91$

$\text{LCM} = \mathbf{1365}$

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(d) LCM of $222, 370$

Using the common division method:

$\begin{array}{c|cc} 2 & 222 \;, & 370 \\ \hline 37 & 111 \;, & 185 \\ \hline & 3 \;, & 5 \end{array}$

$\text{LCM} = 2 \times 37 \times 3 \times 5$

$\text{LCM} = 74 \times 15$

$\text{LCM} = \mathbf{1110}$

(a) Two consecutive even numbers:

Let us take examples: $(2, 4), (10, 12), (24, 26)$.

$\text{HCF}(2, 4) = 2$

$\text{HCF}(10, 12) = 2$

General Statement: The HCF of any two consecutive even numbers is always 2.

Explanation: Consecutive even numbers can be written as $2n$ and $2n+2$. Their difference is 2. Any common factor must also be a factor of their difference. Since both are even, 2 is the highest common factor.

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(b) Two consecutive odd numbers:

Let us take examples: $(3, 5), (15, 17), (21, 23)$.

$\text{HCF}(3, 5) = 1$

$\text{HCF}(21, 23) = 1$

General Statement: The HCF of any two consecutive odd numbers is always 1.

Explanation: The difference between two consecutive odd numbers is 2. The only factors of 2 are 1 and 2. Since odd numbers are not divisible by 2, their only common factor is 1.

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(c) Two even numbers:

Let us take examples: $(4, 8), (12, 18), (20, 30)$.

$\text{HCF}(4, 8) = 4$

$\text{HCF}(12, 18) = 6$

General Statement: The HCF of any two even numbers is always at least 2 (it is always an even number).

Explanation: Since both numbers are even, they are both divisible by 2. Therefore, 2 will always be a common factor.

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(d) Two consecutive numbers:

Let us take examples: $(8, 9), (15, 16), (100, 101)$.

$\text{HCF}(8, 9) = 1$

$\text{HCF}(15, 16) = 1$

General Statement: The HCF of any two consecutive numbers is always 1.

Explanation: Two consecutive numbers do not share any common factors other than 1 because their difference is 1.

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(e) Two co-prime numbers:

Let us take examples: $(4, 9), (8, 15), (7, 10)$.

General Statement: The HCF of two co-prime numbers is always 1.

Explanation: By definition, co-prime numbers are those numbers that have no common factor other than 1.

(a) Examples of such pairs:

1. $(2, 8) \rightarrow \text{LCM} = 8$

2. $(5, 15) \rightarrow \text{LCM} = 15$

3. $(10, 50) \rightarrow \text{LCM} = 50$

4. $(7, 28) \rightarrow \text{LCM} = 28$

5. $(12, 48) \rightarrow \text{LCM} = 48$

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(b) General Statement:

If the larger number is a multiple of the smaller number, then the LCM of the two numbers is the larger number itself.

Algebraic Description:

Let the two numbers be $a$ and $b$.

In such a case:

Given:

Different categories of number pairs: two multiples of $3$, two consecutive even numbers, two consecutive numbers, and two co-prime numbers.

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To Find:

Observations, reasons, and general statements regarding the Least Common Multiple (LCM) for each category.

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Solution:

(a) Two Multiples of $3$

Examples:

(i) $(6, 9)$

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

$\text{LCM}(6, 9) = 3 \times 2 \times 3 = 18$

(ii) $(9, 12)$

$\begin{array}{c|cc} 3 & 9 \;, & 12 \\ \hline & 3 \;, & 4 \end{array}$

$\text{LCM}(9, 12) = 3 \times 3 \times 4 = 36$

(iii) $(12, 18)$

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

$\text{LCM}(12, 18) = 2 \times 3 \times 2 \times 3 = 36$

Observation: The LCM of two multiples of $3$ is also a multiple of $3$.

Reason: Since both numbers are divisible by $3$, their common multiples will also be divisible by $3$. Hence, the LCM must include $3$ as a factor.

General Statement: The LCM of two multiples of $3$ is always a multiple of $3$.

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(b) Two Consecutive Even Numbers

Examples:

(i) $(2, 4)$

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

$\text{LCM}(2, 4) = 2 \times 1 \times 2 = 4$

(ii) $(6, 8)$

$\begin{array}{c|cc} 2 & 6 \;, & 8 \\ \hline & 3 \;, & 4 \end{array}$

$\text{LCM}(6, 8) = 2 \times 3 \times 4 = 24$

(iii) $(10, 12)$

$\begin{array}{c|cc} 2 & 10 \;, & 12 \\ \hline & 5 \;, & 6 \end{array}$

$\text{LCM}(10, 12) = 2 \times 5 \times 6 = 60$

Observation: The LCM of two consecutive even numbers is half of their product.

Reason: Consecutive even numbers always share a common factor of $2$, but not more. Therefore, when finding the LCM, one factor of $2$ overlaps, so the LCM becomes smaller than their product.

General Statement: The LCM of two consecutive even numbers $2n$ and $2n + 2$ is always equal to half of their product.

$\text{LCM}(2n, 2n + 2) = \frac{2n \times (2n + 2)}{2} = n(2n + 2) = 2n^2 + 2n$

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(c) Two Consecutive Numbers

Examples:

(i) $(7, 8)$

$\begin{array}{c|cc} 7 & 7 \;, & 8 \\ \hline 8 & 1 \;, & 8 \\ \hline & 1 \;, & 1 \end{array}$

$\text{LCM}(7, 8) = 7 \times 8 = 56$

(ii) $(9, 10)$

$\begin{array}{c|cc} 9 & 9 \;, & 10 \\ \hline 10 & 1 \;, & 10 \\ \hline & 1 \;, & 1 \end{array}$

$\text{LCM}(9, 10) = 9 \times 10 = 90$

(iii) $(10, 11)$

$\text{LCM}(10, 11) = 110$

Observation: The LCM of two consecutive numbers is equal to their product.

Reason: Consecutive numbers have no common factors other than $1$, so their product is the smallest number divisible by both.

General Statement: The LCM of two consecutive numbers is their product.

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(d) Two Co-prime Numbers

Examples:

(i) $(4, 9)$

$\text{LCM}(4, 9) = 36$

(ii) $(5, 8)$

$\text{LCM}(5, 8) = 40$

(iii) $(7, 10)$

$\text{LCM}(7, 10) = 70$

Observation: The LCM of two co-prime numbers is equal to their product.

Reason: Co-prime numbers do not share any common factors except $1$, so the smallest number that contains both is simply their product.

General Statement: The LCM of two co-prime numbers is equal to their product.

Note: Co-prime numbers are any two natural numbers that have no common factor other than $1$.

Given:

In the first row, the pattern consists of 6 colours: Yellow, Green, Orange, Blue, Pink, and Grey. The blue star is at the 4th position and repeats every 6 stars.

In the second row, the pattern consists of 4 colours: Green, Orange, Yellow, and Blue. The blue star is at the 4th position and repeats every 4 stars.

First meeting point of blue stars is at the 4th position.

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To Find:

The next position where the blue stars in both rows will align again.

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Solution:

To find when the patterns will align again, we need to find the Least Common Multiple (LCM) of the lengths of the two repeating cycles.

Length of the first pattern cycle = 6

Length of the second pattern cycle = 4

We calculate the LCM of 6 and 4:

$\begin{array}{c|cc} 2 & 6 \;, & 4 \\ \hline & 3 \; , & 2 \end{array}$

Since the blue stars first met at the 4th position, they will meet again after every 12 positions.

Verification:

For Row 1 (Pattern of 6): The blue stars appear at positions 4, 10, 16, 22, ...

For Row 2 (Pattern of 4): The blue stars appear at positions 4, 8, 12, 16, 20, ...

Thus, the blue stars will meet next at the 16th position.

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Alternate Method:

We can list the positions of the blue stars in both rows:

Positions in Row 1: $\{4, 10, 16, 22, 28, ...\}$

Positions in Row 2: $\{4, 8, 12, 16, 20, 24, 28, ...\}$

Comparing the two sets, the common positions are 4, 16, 28, and so on.

The next time they meet after the 4th position is the 16th position.

(a) Solution:

Let $A = 5 \times 7 \times 11 \times 11$ and $B = 5 \times 7 \times 7 \times 11 \times 2$.

For $A$ to be a multiple of $B$, all prime factors of $B$ must be present in $A$ with at least the same frequency.

Comparing the factors:

1. $B$ has two $7$s, but $A$ has only one $7$.

2. $B$ has a $2$, but $A$ has no $2$.

Since $B$ contains factors not fully present in $A$, $A$ cannot be divided by $B$ exactly.

Therefore, $5 \times 7 \times 11 \times 11$ is not a multiple of $5 \times 7 \times 7 \times 11 \times 2$.

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(b) Solution:

For $A$ to be a factor of $B$, all prime factors of $A$ must be present in $B$ with at least the same frequency.

Comparing the factors:

1. $A$ has two $11$s ($11 \times 11$).

2. $B$ has only one $11$.

Since $B$ does not have enough $11$s to cover the factors of $A$, $B$ is not divisible by $A$.

Therefore, $5 \times 7 \times 11 \times 11$ is not a factor of $5 \times 7 \times 7 \times 11 \times 2$.

(a) Solution for $3 \times 3 \times 5 \times 7 \times 7$ and $12 \times 7 \times 11$:

First, we express both numbers in their complete prime factor form:

$\text{First number} = 3 \times 3 \times 5 \times 7 \times 7 = 3^2 \times 5 \times 7^2$

$\text{Second number} = 12 \times 7 \times 11 = (2 \times 2 \times 3) \times 7 \times 11 $$ = 2^2 \times 3 \times 7 \times 11$

HCF (Highest Common Factor): We take the common prime factors with the lowest powers.

$\text{HCF} = \mathbf{3 \times 7}$

LCM (Least Common Multiple): We take all prime factors with their highest powers.

$\text{LCM} = \mathbf{2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 7 \times 11}$

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(b) Solution for 45 and 36:

Prime factorisation of 45 and 36:

$\begin{array}{c|cc} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \qquad \begin{array}{c|cc} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$45 = 3 \times 3 \times 5$

$36 = 2 \times 2 \times 3 \times 3$

HCF: Common factors are $3 \times 3$.

$\text{HCF} = \mathbf{3 \times 3}$

LCM: Highest powers of all involved primes ($2^2, 3^2, 5^1$).

$\text{LCM} = \mathbf{2 \times 2 \times 3 \times 3 \times 5}$

Given:

$\text{HCF} = 1$

$\text{LCM} = 66$

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Solution:

When the HCF of two numbers is 1, the numbers are co-prime. For co-prime numbers, the LCM is equal to the product of the two numbers.

$\text{Product of numbers} = \text{HCF} \times \text{LCM} = 1 \times 66 = 66$

We need to find pairs of factors of 66 that are co-prime:

$\begin{array}{c|cc} 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

Possible pairs of factors of $66$ are:

1. $(1, 66) \rightarrow \text{HCF} = 1$

2. $(2, 33) \rightarrow \text{HCF} = 1$

3. $(3, 22) \rightarrow \text{HCF} = 1$

4. $(6, 11) \rightarrow \text{HCF} = 1$

All these pairs satisfy the condition. You can choose any one pair, for example, 6 and 11.

Given:

1. The number of cows is exactly divisible by $3$, $5$, and $7$.

2. The total number of cows is less than $200$.

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To Find:

The total number of cows the cowherd had.

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Solution:

Since the cows passed through $3$, $5$, and $7$ gates in equal numbers, the total number of cows must be a common multiple of $3$, $5$, and $7$.

To find the smallest such number, we calculate the Least Common Multiple (LCM) of $3$, $5$, and $7$.

$\begin{array}{c|cc} 3 & 3 \;, & 5 \;, & 7 \\ \hline 5 & 1 \; , & 5 \; , & 7 \\ \hline 7 & 1 \; , & 1 \; , & 7 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

$\text{LCM} = 3 \times 5 \times 7 = 105$

The total number of cows must be a multiple of $105$. The multiples of $105$ are:

$105 \times 1 = 105$

$105 \times 2 = 210$

Since it is given that the cowherd had less than $200$ cows, only $105$ satisfies the condition.

Therefore, the cowherd had $105$ cows.

Given:

Dimensions of the box: $12\text{ cm} \times 18\text{ cm} \times 36\text{ cm}$.

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To Find:

The size of the cubes that can be packed without leaving gaps.

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Solution:

For a cube to be packed without leaving any gaps, its side length must be a common factor of the length, width, and height of the box.

We check each option for divisibility against $12, 18, \text{ and } 36$:

(a) $9\text{ cm}$: $12$ is not divisible by $9$. (Not possible)

(b) $6\text{ cm}$: $12$, $18$, and $36$ are all divisible by $6$. (Possible)

(c) $4\text{ cm}$: $18$ is not divisible by $4$. (Not possible)

(d) $3\text{ cm}$: $12$, $18$, and $36$ are all divisible by $3$. (Possible)

(e) $2\text{ cm}$: $12$, $18$, and $36$ are all divisible by $2$. (Possible)

Therefore, the cubes of sizes $6\text{ cm}$, $3\text{ cm}$, and $2\text{ cm}$ can be packed without leaving gaps.

Given:

The two numbers are $306$ and $36$.

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To Find:

The largest number that perfectly divides both $306$ and $36$. In mathematical terms, we need to find the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of $306$ and $36$.

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Solution:

To find the HCF, we will first perform the prime factorisation of both numbers.

Prime factorisation of $306$:

$\begin{array}{c|cc} 2 & 306 \\ \hline 3 & 153 \\ \hline 3 & 51 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$

Therefore, we can write:

Prime factorisation of $36$:

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

Therefore, we can write:

Now, we identify the common prime factors in both expressions:

The common factors are one $2$ and two $3$s ($3 \times 3$).

Thus, the largest number that perfectly divides both $306$ and $36$ is $18$.

Comparing this with the given options, the correct choice is (c).

Given:

1. The number must be perfectly divisible by $3, 4, 5,$ and $7$.

2. The number must leave a remainder of $10$ when divided by $11$.

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To Find:

The smallest number satisfying both conditions.

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Solution:

First, we need to find the smallest number divisible by $3, 4, 5,$ and $7$. This number is the Least Common Multiple (LCM) of these four numbers.

Since $3, 4, 5,$ and $7$ are co-prime (they have no common factors other than 1), their LCM is simply their product.

$\begin{array}{c|cccc} 2 & 3 \;, & 4 \;, & 5 \;, & 7 \\ \hline 2 & 3 \;, & 2 \;, & 5 \;, & 7 \\ \hline & 3 \;, & 1 \;, & 5 \;, & 7 \end{array}$

Any number divisible by $3, 4, 5,$ and $7$ must be a multiple of $420$. Let the required number be $420k$, where $k$ is a positive integer.

According to the second condition, when $420k$ is divided by $11$, the remainder should be $10$. We can express $420$ in terms of a multiple of $11$ as follows:

Substituting this into our expression $420k$:

$420k = (11 \times 38 + 2)k = 11 \times 38k + 2k$

For $420k$ to leave a remainder of $10$ when divided by $11$, the term $2k$ must leave a remainder of $10$ when divided by $11$.

We test values of $k$ starting from $1$:

If $k = 1$, $2k = 2$ (Remainder is 2)

If $k = 2$, $2k = 4$ (Remainder is 4)

If $k = 3$, $2k = 6$ (Remainder is 6)

If $k = 4$, $2k = 8$ (Remainder is 8)

If $k = 5$, $2k = 10$ (Remainder is 10)

The smallest value for $k$ that satisfies the condition is $k = 5$.

Now, we find the required number by multiplying $420$ by $5$:

Given:

1. When the number $6$ was called, no one got out. This means the total number of children ($N$) is exactly divisible by $6$.

2. When the number $9$ was called, no one got out. This means $N$ is exactly divisible by $9$.

3. When the number $10$ was called, some people got out. This means $N$ is not exactly divisible by $10$.

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To Find:

The possible value of $N$ from the given options.

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Solution:

Since the number of children $N$ is divisible by both $6$ and $9$, it must be a multiple of the Least Common Multiple (LCM) of $6$ and $9$.

Finding LCM of $6$ and $9$:

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

This means the total number of children $N$ must be a multiple of $18$. The multiples of $18$ are:

$18, 36, 54, 72, 90, 108, \dots$

Now, we check the given options based on two conditions: (i) it must be a multiple of $18$, and (ii) it must not be a multiple of $10$ (because some people got out when $10$ was called).

Evaluating the options:

1. Option (a) 72: $72 = 18 \times 4$. It is a multiple of $18$. It is not divisible by $10$. This satisfies all conditions.

2. Option (b) 90: $90 = 18 \times 5$. It is a multiple of $18$. However, $90$ is also divisible by $10$. If there were $90$ children, no one would have gotten out when $10$ was called. So, this is incorrect.

3. Option (c) 45: $45$ is not divisible by $18$. Incorrect.

4. Option (d) 3: $3$ is not divisible by $18$. Incorrect.

5. Option (e) 36: $36 = 18 \times 2$. It is a multiple of $18$. It is not divisible by $10$. This also satisfies all conditions.

Since both $72$ and $36$ are multiples of $18$ and not divisible by $10$, both are mathematically correct possibilities. However, in most standard contexts for this problem, we look for the choices provided.

The correct choices from the list are (a) $72$ and (e) $36$.

Given:

Two different prime numbers $m$ and $n$.

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Solution:

Prime numbers are those numbers that have only two factors, $1$ and the number itself. Since $m$ and $n$ are different prime numbers, they are co-prime to each other (their $\text{HCF} = 1$).

For any two co-prime numbers, the Least Common Multiple (LCM) is the product of the two numbers.

Since prime numbers are greater than or equal to $2$, their product $m \times n$ will always be greater than both $m$ and $n$.

Therefore, the correct statement is:

(c) Greater than both numbers

Given:

Initial head start of the rabbit $= 150\text{ feet}$.

Distance covered by the dog in one leap $= 9\text{ feet}$.

Distance covered by the rabbit in one leap $= 7\text{ feet}$.

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To Find:

The number of leaps required for the dog to catch the rabbit.

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Solution:

For every leap taken simultaneously, the dog gains some distance over the rabbit. This is called the relative speed or relative gain per leap.

To catch the rabbit, the dog needs to cover the entire head start of $150$ feet using this relative gain.

$\text{Number of leaps} = \frac{\text{Total head start}}{\text{Relative gain per leap}}$

Therefore, the dog will catch up with the rabbit in 75 leaps.

To Find:

The smallest number that is a multiple of $1, 2, 3, 4, 5, 6, 8, 9, \text{ and } 10$. This number is the Least Common Multiple (LCM) of the given set.

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Solution:

We use the common division method to find the LCM of the numbers:

$\begin{array}{c|cc} 2 & 1 \;, & 2 \;, & 3 \;, & 4 \;, & 5 \;, & 6 \;, & 8 \;, & 9 \;, & 10 \\ \hline 2 & 1 \;, & 1 \;, & 3 \;, & 2 \;, & 5 \;, & 3 \;, & 4 \;, & 9 \;, & 5 \\ \hline 2 & 1 \;, & 1 \;, & 3 \;, & 1 \;, & 5 \;, & 3 \;, & 2 \;, & 9 \;, & 5 \\ \hline 3 & 1 \;, & 1 \;, & 3 \;, & 1 \;, & 5 \;, & 3 \;, & 1 \;, & 9 \;, & 5 \\ \hline 3 & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 5 \;, & 1 \;, & 1 \;, & 3 \;, & 5 \\ \hline 5 & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 1 \end{array}$

Now, we multiply all the prime divisors:

$\text{LCM} = 2 \times 2 \times 2 \times 3 \times 3 \times 5$

$\text{LCM} = 8 \times 9 \times 5$

$\text{LCM} = 72 \times 5 = 360$

Therefore, the smallest number that is a multiple of all these numbers is 360.

Note: In Grade 6, we learned that $2520$ is the smallest number divisible by all numbers from $1$ to $10$. However, since $7$ is missing from the list in this question, the answer is $360$.

Given Expression:

$\frac{8}{15} + \frac{1}{20} + \frac{7}{36} + \frac{11}{63} + \frac{1}{21}$

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Solution:

To add these fractions efficiently, we first find the Least Common Multiple (LCM) of the denominators: $15, 20, 36, 63, \text{ and } 21$.

$\begin{array}{c|cc} 2 & 15 \;, & 20 \;, & 36 \;, & 63 \;, & 21 \\ \hline 2 & 15 \;, & 10 \;, & 18 \;, & 63 \;, & 21 \\ \hline 3 & 15 \;, & 5 \;, & 9 \;, & 63 \;, & 7 \\ \hline 3 & 5 \;, & 5 \;, & 3 \;, & 21 \;, & 7 \\ \hline 5 & 5 \;, & 5 \;, & 1 \;, & 7 \;, & 7 \\ \hline 7 & 1 \;, & 1 \;, & 1 \;, & 7 \;, & 7 \\ \hline & 1 \;, & 1 \;, & 1 \;, & 1 \;, & 1 \end{array}$

$\text{LCM} = 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 4 \times 9 \times 35 = 1260$

Now, convert each fraction to an equivalent fraction with denominator $1260$:

$\frac{8}{15} = \frac{8 \times 84}{1260} = \frac{672}{1260}$

$\frac{1}{20} = \frac{1 \times 63}{1260} = \frac{63}{1260}$

$\frac{7}{36} = \frac{7 \times 35}{1260} = \frac{245}{1260}$

$\frac{11}{63} = \frac{11 \times 20}{1260} = \frac{220}{1260}$

$\frac{1}{21} = \frac{1 \times 60}{1260} = \frac{60}{1260}$

Adding the numerators:

$\begin{array}{cc} & 6 & 7 & 2 \\ & & 6 & 3 \\ & 2 & 4 & 5 \\ & 2 & 2 & 0 \\ + & & 6 & 0 \\ \hline 1 & 2 & 6 & 0 \\ \hline \end{array}$

The total sum is:

$\text{Sum} = \frac{1260}{1260}$

On cancelling the numerator and denominator:

$\text{Sum} = \frac{\cancel{1260}^{1}}{\cancel{1260}_{1}} = 1$

Therefore, the final result is 1.