| Content On This Page | ||
|---|---|---|
| Word Problems on Large Numbers | Uses of Fractions, Decimals, and Rational Numbers in Applications | Conversion of Units |
| Playing with Numbers: Puzzles and Digit Manipulation | ||
Numerical Applications and Word Problems
Word Problems on Large Numbers
Dealing with Large Numbers in Context
Word problems provide realistic scenarios where we apply mathematical concepts and operations. Problems involving large numbers are prevalent in fields such as population studies (demography), economics (financial figures), astronomy (distances, masses), environmental science, and big data analysis. Solving these problems requires not only identifying the correct operations but also being able to read, write, and perform arithmetic accurately with numbers containing many digits.
The process of solving a word problem typically involves several steps:
- Understand the Problem: Read the problem carefully to grasp the context. Identify what information is given (the numbers and their meanings) and what needs to be found. Sometimes, writing down the given information and the question separately helps clarify the problem.
- Plan the Solution: Determine which arithmetic operation(s) are needed to solve the problem. Based on the question (e.g., "how much total," "how much more," "how much less," "how many times," "how many in each group"), decide if you need to add, subtract, multiply, or divide. Break down multi-step problems into a sequence of simpler operations.
- Perform the Calculations: Carry out the arithmetic operations using the given large numbers. Pay close attention to place values. Using the appropriate numeration system (Indian or International, as used in the problem statement) with commas can help in reading and writing the numbers correctly. Column addition, subtraction, and long multiplication/division are standard methods for handling multi-digit numbers.
- Verify and State the Answer: Check if your calculation is correct. One way to do this is by estimating the answer beforehand and comparing it to your calculated result. Finally, write the answer in a clear sentence that directly addresses the question asked in the word problem, including the correct units (e.g., $\textsf{₹}$, tonnes, kg, km, people).
Examples of Word Problems Involving Large Numbers
Example 1. In the year $2023$, Country A produced $1,45,78,65,000 \text{ tonnes}$ of food grains. Country B produced $1,29,50,40,750 \text{ tonnes}$ in the same year. How much more food grains did Country A produce than Country B?
Answer:
1. Understand the Problem:
Given: Production by Country A ($1,45,78,65,000$ tonnes) and Production by Country B ($1,29,50,40,750$ tonnes). We need to find the difference between these two quantities.
2. Plan the Solution:
To find "how much more", we need to subtract the smaller quantity (Country B's production) from the larger quantity (Country A's production).
Operation needed: Subtraction.
3. Perform the Calculations:
We need to calculate $1,45,78,65,000 - 1,29,50,40,750$. Align the numbers vertically by place value and perform column subtraction, using borrowing where necessary.
Let's perform the subtraction carefully from right to left:
- Ones place: $0 - 0 = 0$.
- Tens place: $0 - 5$. Need to borrow from the left. The Tens place (0) borrows from the Hundreds place (0), which borrows from the Thousands place (5). Thousands digit 5 becomes 4. Hundreds digit 0 becomes 10. Hundreds digit 10 lends 1 to the Tens place, becoming 9. Tens digit 0 becomes 10. $10 - 5 = 5$.
- Hundreds place: $9 - 7 = 2$ (This digit was 0, borrowed from, now is 9).
- Thousands place: $4 - 0 = 4$ (This digit was 5, lent to the right, now is 4).
- Ten Thousands place: $6 - 4 = 2$.
- Lakhs place: $8 - 0 = 8$.
- Ten Lakhs place: $7 - 5 = 2$.
- Crores place: $5 - 9$. Need to borrow from the Ten Crores place (4). Ten Crores digit 4 becomes 3. Crores digit 5 becomes 15. $15 - 9 = 6$.
- Ten Crores place: $3 - 2 = 1$ (This digit was 4, lent to the right, now is 3).
- Arabs place: $1 - 1 = 0$.
The result of the subtraction is $016,28,24,250$, which is written as $16,28,24,250$ (omitting the leading zero).
4. Verify and State the Answer:
Estimate: Country A produced about $145$ crore tonnes. Country B produced about $129$ crore tonnes. Difference $\approx 145 - 129 \approx 16$ crore tonnes. Our calculated answer is $16,28,24,250$, which is about 16 crore. The answer is reasonable.
The question asks how much more Country A produced than Country B.
Answer: Country A produced $\mathbf{16,28,24,250 \text{ tonnes}}$ more food grains than Country B.
Example 2. The annual production of sugar from a factory is $12,50,000 \text{ kg}$. If the factory operates for $320$ days in a year, what is the average daily production of sugar in kilograms?
Answer:
1. Understand the Problem:
Given: Total annual production ($12,50,000 \text{ kg}$) and number of operating days ($320 \text{ days}$). We need to find the average amount produced per day.
2. Plan the Solution:
Average daily production is found by dividing the total production by the number of days. Operation needed: Division.
3. Perform the Calculations:
We need to calculate $\frac{12,50,000}{320}$. We can simplify the division by canceling a common factor of 10 from the numerator and the denominator (removing one zero from each):
Now, perform the long division of $1,25,000$ by $32$:
$$ \begin{array}{r} 3906.25 \\ 32{\overline{\smash{\big)}\,125000.00}} \\ \underline{-96\downarrow\phantom{00.00}}\\ 290\phantom{0.00}\\ \underline{-288\downarrow\phantom{0.00}}\\ 20\phantom{0.0}\\ \underline{-\phantom{0}0\downarrow\phantom{0.0}}\\ 200\phantom{.0}\\ \underline{-192\downarrow\phantom{.0}}\\ 80\phantom{.}\\ \underline{-64\downarrow}\\ 160\\ \underline{-160}\\ 0 \end{array} $$Detailed steps of the long division:
- Divide 125 by 32: $125 \div 32 = 3$ with remainder $125 - (32 \times 3) = 125 - 96 = 29$. Write 3 in the quotient.
- Bring down the next digit (0) to form 290. Divide 290 by 32: $290 \div 32 = 9$ with remainder $290 - (32 \times 9) = 290 - 288 = 2$. Write 9 in the quotient.
- Bring down the next digit (0) to form 20. Divide 20 by 32: $20 \div 32 = 0$ with remainder $20 - (32 \times 0) = 20$. Write 0 in the quotient.
- Bring down the next digit (0) to form 200. Divide 200 by 32: $200 \div 32 = 6$ with remainder $200 - (32 \times 6) = 200 - 192 = 8$. Write 6 in the quotient. We have used all digits before the decimal point in the original number, so place a decimal point in the quotient.
- Bring down the next digit (0) from the decimal part to form 80. Divide 80 by 32: $80 \div 32 = 2$ with remainder $80 - (32 \times 2) = 80 - 64 = 16$. Write 2 after the decimal point in the quotient.
- Bring down the next digit (0) to form 160. Divide 160 by 32: $160 \div 32 = 5$ with remainder $160 - (32 \times 5) = 160 - 160 = 0$. Write 5 in the quotient. The remainder is 0, so the division terminates.
The quotient is $3906.25$.
4. Verify and State the Answer:
Estimate: Total production $\approx 12,00,000$ kg. Days $\approx 300$. Average daily production $\approx 12,00,000 \div 300 = 12000 \div 3 = 4000$ kg. Our calculated answer $3906.25$ is close to the estimate $4000$, which is reasonable.
The question asks for the average daily production in kilograms.
Answer: The average daily production of sugar from the factory is $\mathbf{3906.25 \text{ kg}}$.
Solving word problems involving large numbers requires careful reading, planning, and accurate execution of arithmetic operations, along with the ability to interpret the numbers in the context of the problem.
Uses of Fractions, Decimals, and Rational Numbers in Applications
Fractions, decimals, and rational numbers are fundamental components of the number system that enable us to represent and work with quantities that are not restricted to whole units. They are essential tools in countless applications across everyday life, various professions, and scientific fields. Since integers, fractions, and terminating or repeating decimals are all types of rational numbers, their applications collectively represent the applications of rational numbers in a broad sense.
Applications of Fractions
Fractions ($\frac{p}{q}$) are particularly useful for representing parts of a whole, proportions, ratios, and the result of exact divisions.
- Measurement: Expressing measurements that are parts of standard units. Common examples include measurements of length, weight, volume, and time.
Example: $\frac{1}{2}$ kg of vegetables, $\frac{3}{4}$ cup of flour in a recipe, $\frac{1}{4}$ of an hour, $\frac{1}{8}$ of an inch.
- Sharing and Distribution: Dividing items, resources, or quantities equally among a number of people or entities.
Example: Sharing a pizza or cake into equal slices (e.g., $\frac{1}{8}$ths), dividing assets among $\frac{1}{3}$ shareholders of a company, splitting a land parcel into $\frac{1}{4}$ shares.
- Recipes and Cooking/Baking: Recipes frequently specify ingredient amounts using fractions.
Example: Add $\frac{1}{2}$ teaspoon of salt, use $\frac{3}{4}$ cup of milk, bake for $1\frac{1}{2}$ hours.
- Construction and Carpentry: Working with dimensions, plans, and cutting materials often involves fractions of standard units of length, especially in systems using feet and inches.
- Expressing Ratios and Rates: Comparing two quantities as a ratio (e.g., a mixing ratio of cement to sand might be $1:3$ or $\frac{1}{3}$), or expressing rates of change (e.g., speed is distance/time, density is mass/volume).
- Probability: Calculating and expressing the likelihood of events as a ratio of the number of favorable outcomes to the total number of possible outcomes. Probabilities are always numbers between 0 and 1, often expressed as fractions or decimals.
Example: The probability of rolling a 4 on a fair six-sided die is $\frac{1}{6}$.
- Music: Musical notation uses fractions to represent the relative duration of notes (e.g., a half note ($\frac{1}{2}$), a quarter note ($\frac{1}{4}$), an eighth note ($\frac{1}{8}$)).
- Original length of ribbon = $\frac{5}{6} \text{ metres}$
- Length cut off = $\frac{1}{4} \text{ metres}$
- Money and Finance: All monetary systems are based on decimals. Prices, costs, taxes, interest rates, bank balances, and financial calculations use decimals.
Example: $\textsf{₹} 50.75$, a product costing $\textsf{₹} 499.99$, calculating $18\%$ GST on a price, calculating simple or compound interest.
- Scientific and Engineering Measurements: Most measurements in science and engineering fields (physics, chemistry, biology, mechanical engineering, electrical engineering, etc.) are recorded and used in decimal form for precision and ease of calculation.
Example: Body temperature $37.5^\circ \text{C}$, speed $9.8 \text{ m/s}^2$, mass $1.05 \text{ kg}$, current $2.5 \text{ A}$, voltage $230 \text{ V}$.
- Computer Science: While computers use binary internally, numbers are often input, displayed, and processed in decimal format. Floating-point numbers in programming languages approximate real numbers using a decimal representation system.
- Data Analysis and Statistics: Averages, percentages, proportions, and various statistical measures (like standard deviation, correlation coefficients) are frequently presented as decimals.
- Sports: Timings in races (e.g., 10.5 seconds), scores in various events (e.g., diving, gymnastics), calculating averages (like batting averages in cricket/baseball, points per game in basketball) use decimals.
- Speed = $60.5 \text{ km/h}$
- Time = $3.5 \text{ hours}$
- $60.5$ has 1 decimal place (the digit 5).
- $3.5$ has 1 decimal place (the digit 5).
- Counting Discrete Objects: Using integers (a subset of rational numbers) to count people, objects, money, etc.
- Representing Quantities: Using fractions and decimals to represent quantities that are not whole numbers, covering measurements, proportions, etc.
- Comparison and Ordering: Rational numbers can be ordered on the number line, allowing comparison of quantities in various applications.
- Solving Equations: Many real-world problems translate into algebraic equations (especially linear equations) whose solutions are rational numbers.
- Proportions and Scaling: Used extensively in scaling recipes, maps, blueprints, architectural designs, and models while maintaining correct proportions.
- Computer Representations: Rational numbers (including integers and terminating decimals) can often be represented exactly in computer systems (unlike most irrational numbers which can only be approximated).
- Ratios and Rates: Expressing relationships between quantities as ratios or rates, which are inherently rational concepts.
Example 1. A piece of ribbon is $\frac{5}{6} \text{ metres}$ long. If you cut off $\frac{1}{4}$ of a metre, how much ribbon is left?
Answer:
Given:
To Find: Length of ribbon left.
Solution:
To find the length of ribbon remaining, subtract the length that was cut off from the original length.
$\text{Length left} = \text{Original length} - \text{Length cut off}$
$\quad = \frac{5}{6} - \frac{1}{4}$
To subtract fractions with different denominators, find a common denominator. The denominators are 6 and 4. The Least Common Multiple (LCM) of 6 and 4 is 12. Convert both fractions to equivalent fractions with a denominator of 12.
$\quad \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$
[Multiplying numerator and denominator by 2]
$\quad \frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$
[Multiplying numerator and denominator by 3]
Now subtract the like fractions:
$\quad \frac{10}{12} - \frac{3}{12} = \frac{10 - 3}{12} = \frac{7}{12}$
The result is $\frac{7}{12}$. The units are metres.
Answer: $\mathbf{\frac{7}{12} \text{ metres}}$ of ribbon is left.
Applications of Decimals
Decimal numbers are essentially fractions with denominators as powers of 10. They are widely used because they are convenient for calculations, especially with calculators and computers, and are compatible with the base-10 number system.
Example 1. A car travels at a speed of $60.5 \text{ km/h}$. How much distance will it cover in $3.5$ hours?
Answer:
Given:
To Find: Distance covered.
Solution:
The relationship between distance, speed, and time is Distance = Speed $\times$ Time.
$\text{Distance} = 60.5 \text{ km/h} \times 3.5 \text{ hours}$
We need to multiply the decimal numbers 60.5 and 3.5. Multiply them as if they were whole numbers (605 $\times$ 35), then place the decimal point in the product.
$$ \begin{array}{cc} & 605 \\ \times & 35 \\ \hline & 3025 & \leftarrow 605 \times 5 \\ 1815 \times & \leftarrow 605 \times 30 \\ \hline 21175 \\ \hline \end{array} $$Count the total number of decimal places in the numbers being multiplied:
The total number of decimal places in the factors is $1 + 1 = 2$.
Place the decimal point in the product (21175) such that there are 2 decimal places, counting from the rightmost digit.
$$ 211.75 $$The result is 211.75. The units are km.
Answer: The car will cover a distance of $\mathbf{211.75 \text{ km}}$.
Applications of Rational Numbers (Integers, Fractions, Decimals)
Since rational numbers ($\mathbb{Q}$) include all integers, fractions, and terminating or repeating decimals, their applications encompass all the uses mentioned above for integers, fractions, and decimals. Rational numbers are used in virtually every context where quantities can be counted precisely or expressed as exact ratios or finite parts of a whole.
Rational numbers are fundamental for describing countable or precisely measurable quantities and analyzing their relationships and changes.
Conversion of Units
The Importance of Unit Conversion
In practical applications, measurements of physical quantities (such as length, mass, time, temperature, volume, speed, etc.) are often expressed using different units. For example, length can be measured in metres, centimeters, kilometers, feet, or inches. To perform calculations involving these measurements, compare values accurately, or combine quantities, it is essential that all values are expressed in the same, consistent unit. This process is called unit conversion.
Unit conversion is based on established equivalences or relationships between different units of the same quantity. These relationships are stated as equalities (e.g., $1 \text{ kilogram} = 1000 \text{ grams}$, $1 \text{ hour} = 60 \text{ minutes}$, $1 \text{ metre} = 100 \text{ centimetres}$).
Using Conversion Factors
Unit conversion is performed by multiplying the measurement by one or more conversion factors. A conversion factor is a ratio of two equivalent quantities expressed in different units. Because the numerator and the denominator of a conversion factor represent the same amount, the ratio is equal to $1$. Multiplying a measurement by a conversion factor is essentially multiplying it by $1$, so the value of the measurement remains unchanged, but its units and numerical representation change.
Example: The relationship $1 \text{ metre} = 100 \text{ centimetres}$ can be written as the equality $1 \text{ m} = 100 \text{ cm}$. From this equality, we can form two conversion factors:
$\quad \frac{100 \text{ cm}}{1 \text{ m}} = 1 \quad \text{and} \quad \frac{1 \text{ m}}{100 \text{ cm}} = 1$
To convert a quantity from 'Unit A' to 'Unit B', you multiply the quantity in 'Unit A' by the conversion factor that has 'Unit A' in the denominator and 'Unit B' in the numerator. This allows the original unit ('Unit A') to cancel out, leaving the desired unit ('Unit B').
$\text{Quantity in Unit B} = (\text{Quantity in Unit A}) \times \left(\frac{\text{Equivalent amount in Unit B}}{\text{Equivalent amount in Unit A}}\right)$
Examples of Simple Unit Conversion
Example 1. Convert $2.5$ kilograms to grams.
Answer:
The given quantity is $2.5$ kg. We want to convert it to grams (g).
We know the relationship between kilograms and grams: $1 \text{ kilogram (kg)} = 1000 \text{ grams (g)}$.
We need a conversion factor that cancels out kilograms (kg) and introduces grams (g). The conversion factor is $\frac{1000 \text{ g}}{1 \text{ kg}}$.
Multiply the given quantity by the conversion factor:
$\quad 2.5 \text{ kg} = 2.5 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}}$
Cancel the units 'kg' and perform the multiplication:
$\quad = 2.5 \times 1000 \text{ g} = 2500 \text{ g}$
Answer: $2.5$ kilograms is equal to $\mathbf{2500 \text{ grams}}$.
Example 2. How many hours are there in $3$ days and $12$ hours?
Answer:
The quantity is given in mixed units (days and hours). We want the total time in hours. The 12 hours are already in the desired unit. We need to convert the 3 days to hours and then add the 12 hours.
We know the relationship: $1 \text{ day} = 24 \text{ hours}$.
To convert days to hours, we need a conversion factor with 'day' in the denominator and 'hours' in the numerator: $\frac{24 \text{ hours}}{1 \text{ day}}$.
Convert 3 days to hours:
$\quad 3 \text{ days} = 3 \text{ days} \times \frac{24 \text{ hours}}{1 \text{ day}}$
Cancel units and multiply:
$\quad = 3 \times 24 \text{ hours} = 72 \text{ hours}$
Now, add the extra 12 hours:
$\quad \text{Total hours} = 72 \text{ hours} + 12 \text{ hours} = 84 \text{ hours}$
Answer: There are $\mathbf{84 \text{ hours}}$ in $3$ days and $12$ hours.
Compound Unit Conversions
Some conversions involve units in both the numerator and the denominator, such as converting speed (distance/time) or density (mass/volume). In such cases, you apply multiple conversion factors sequentially.
Example 1. Convert a speed of $36 \text{ km/h}$ to metres per second (m/s).
Answer:
We need to convert the unit of distance from kilometers (km) to metres (m) and the unit of time from hours (h) to seconds (s).
Known relationships:
- $1 \text{ km} = 1000 \text{ m}$
- $1 \text{ hour} = 60 \text{ minutes}$
- $1 \text{ minute} = 60 \text{ seconds}$
- So, $1 \text{ hour} = 60 \text{ minutes} \times \frac{60 \text{ seconds}}{1 \text{ minute}} = 3600 \text{ seconds}$.
We start with $36 \frac{\text{km}}{\text{h}}$. We need conversion factors that will cancel km and introduce m, and cancel h and introduce s.
- To convert km to m, use the conversion factor $\frac{1000 \text{ m}}{1 \text{ km}}$. (km is in the denominator to cancel the km in the numerator of km/h).
- To convert h to s, use the conversion factor $\frac{1 \text{ hour}}{3600 \text{ seconds}}$. (hour must be in the numerator to cancel the hour in the denominator of km/h).
Multiply the given speed by these two conversion factors:
$\quad 36 \frac{\text{km}}{\text{h}} = 36 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}}$
Cancel the units 'km' and 'hour'. Multiply and divide the numerical values:
$\quad = 36 \times \frac{1000}{3600} \frac{\text{m}}{\text{s}}$
Simplify the fraction $\frac{1000}{3600}$ (cancel zeros, divide by 10): $\frac{1000}{3600} = \frac{10}{36}$. Further simplify by dividing by 2: $\frac{10}{36} = \frac{5}{18}$. Or cancel 1000 from numerator and denominator: $\frac{\cancel{1000}^{1}}{\cancel{3600}_{3.6}}$ then $36/3.6=10$. Let's simplify $\frac{1000}{3600}$ by dividing both by 100, then by 10, then by 2: $\frac{10}{36}=\frac{5}{18}$. So, $36 \times \frac{5}{18}$.
$\quad = 36 \times \frac{5}{18} \frac{\text{m}}{\text{s}}$
Cancel 18 into 36:
$\quad = \cancel{36}^2 \times \frac{5}{\cancel{18}_1} \frac{\text{m}}{\text{s}} = 2 \times 5 \frac{\text{m}}{\text{s}}$
$\quad = 10 \frac{\text{m}}{\text{s}}$
Answer: A speed of $36 \text{ km/h}$ is equal to $\mathbf{10 \text{ m/s}}$.
Unit conversion is a fundamental skill required whenever measurements are used in calculations or comparisons, ensuring consistency and accuracy in applied mathematics.
Playing with Numbers: Puzzles and Digit Manipulation
Exploring Number Properties and Structures
Mathematics is not just about calculations; it also involves exploring the fascinating properties and structures of numbers. "Playing with numbers" often involves engaging in puzzles, games, and investigations that reveal interesting patterns and relationships between integers and their digits. These activities encourage logical thinking, problem-solving skills, and a deeper appreciation for numbers.
This often involves representing numbers in a general form using variables for digits, applying concepts of place value, and setting up algebraic equations to model the relationships described in word problems about digits.
Number Puzzles Based on Digits
Many common number puzzles involve manipulating the digits of an integer, such as reversing the digits, swapping digits, summing the digits, or inserting digits, and then establishing relationships between the original number and the number(s) created through these manipulations. Using the general form of numbers is crucial for solving these types of problems algebraically.
Recall the general form of two-digit and three-digit numbers based on their digits:
- A two-digit number with tens digit $a$ and ones digit $b$ can be written as $10a + b$. Note that $a$ is a digit from 1 to 9 ($a \in \{1, 2, \ldots, 9\}$), and $b$ is a digit from 0 to 9 ($b \in \{0, 1, \ldots, 9\}$). The number formed by reversing the digits is $10b + a$.
- A three-digit number with hundreds digit $a$, tens digit $b$, and ones digit $c$ can be written as $100a + 10b + c$. Note that $a \in \{1, 2, \ldots, 9\}$, and $b, c \in \{0, 1, \ldots, 9\}$. The number formed by reversing the digits is $100c + 10b + a$.
Using these general forms, we can translate word problems about digits into algebraic equations and solve them.
Example 1. The sum of the digits of a two-digit number is $8$. The number obtained by interchanging the digits is $18$ more than the original number. Find the original number.
Answer:
Let:
- The tens digit of the original number be $a$.
- The ones digit of the original number be $b$.
The original number can be represented in general form as $10a + b$. The digits must satisfy $a \in \{1, 2, \ldots, 9\}$ and $b \in \{0, 1, \ldots, 9\}$.
The number obtained by interchanging (reversing) the digits has $b$ as the tens digit and $a$ as the ones digit. Its general form is $10b + a$.
Translate the given conditions into equations:
1. "The sum of the digits of a two-digit number is $8$."
The sum of the tens digit and the ones digit is 8.
$\quad a + b = 8$
... (i)
2. "The number obtained by interchanging the digits is $18$ more than the original number."
The interchanged number is equal to the original number plus 18.
$\quad \text{Interchanged number} = \text{Original number} + 18$
$\quad 10b + a = (10a + b) + 18$
... (ii)
Solve the system of equations:
We have two equations with two variables $a$ and $b$:
Equation (i): $a + b = 8$
Equation (ii): $10b + a = 10a + b + 18$. Let's simplify Equation (ii).
Subtract $10a$ and $b$ from both sides of Equation (ii):
$\quad 10b - b + a - 10a = 18$
$\quad 9b - 9a = 18$
Divide the entire equation by 9:
$\quad b - a = 2$
... (iii)
Now we solve the system of linear equations formed by Equation (i) and Equation (iii):
Equation (i): $a + b = 8$}
Equation (iii): $-a + b = 2$}
We can add Equation (i) and Equation (iii) to eliminate $a$:
$\quad (a + b) + (-a + b) = 8 + 2$
$\quad a + b - a + b = 10$
$\quad 2b = 10$
Divide by 2:
$\quad b = 5$
Substitute the value of $b$ into Equation (i):
$\quad a + 5 = 8$
$\quad a = 8 - 5 = 3$
The digits are $a=3$ and $b=5$. The original number is $10a + b = 10(3) + 5 = 30 + 5 = 35$.
Check: Original number is 35. Sum of digits $3+5=8$ (Condition 1 satisfied). Interchanged number is 53. Is 53 eighteen more than 35? $53 - 35 = 18$. Yes, 53 is 18 more than 35 (Condition 2 satisfied). The digits $a=3, b=5$ are valid ($a \neq 0$).
Answer: The original number is $\mathbf{35}$.
Pattern Recognition in Sequences and Operations
Another enjoyable aspect of playing with numbers involves identifying patterns in sequences of numbers or the results of specific operations. Once a pattern is recognized, you can often predict future terms or derive a rule that governs the pattern.
Example 1. Observe the pattern formed by squaring numbers consisting only of the digit 1: $$ 1^2 = 1 $$ $$ 11^2 = 121 $$ $$ 111^2 = 12321 $$ $$ 1111^2 = 1234321 $$ Write the next two steps based on the observed pattern.
Answer:
Let's analyze the given steps:
- $1^2$: The base has one digit '1'. The result is 1.
- $11^2$: The base has two digits '1'. The result is 121. The digits ascend from 1 to 2, then descend back to 1.
- $111^2$: The base has three digits '1'. The result is 12321. The digits ascend from 1 to 3, then descend back to 1.
- $1111^2$: The base has four digits '1'. The result is 1234321. The digits ascend from 1 to 4, then descend back to 1.
The pattern is clear: when squaring a number consisting of $n$ digits of '1', the result is a number formed by ascending digits from 1 to $n$ and then descending back down to 1.
Following this pattern for the next two steps:
Step 5: The base number will have $n=5$ digits of '1', which is 11111. The result will ascend from 1 to 5 and then descend back to 1.
$\quad 11111^2 = 123454321$
Step 6: The base number will have $n=6$ digits of '1', which is 111111. The result will ascend from 1 to 6 and then descend back to 1.
$\quad 111111^2 = 12345654321$
These next two steps fit the observed pattern.
These types of number puzzles highlight the structure and patterns within numbers and operations, encouraging logical reasoning, observation, and the ability to generalize mathematical ideas.