NCERT Class 8 Maths: Playing With Numbers

Understanding the Generalised Form of Numbers

In earlier classes, we wrote numbers in expanded form. For example, $52$ is written as $50 + 2$, which is $10 \times 5 + 2$. Similarly, a three-digit number like $357$ is $300 + 50 + 7$, which is $100 \times 3 + 10 \times 5 + 7$. In general, we can represent any two-digit number with tens digit '$a

This generalised notation helps us prove interesting properties. For instance, if you add a two-digit number $ab$ and its reversed form $ba$, you get $(10a + b) + (10b + a) = 11a + 11b = 11(a + b)$. This shows that the sum of any two-digit number and its reverse is always a multiple of 11!

Step-by-Step Guide to Solving Cryptarithms (Letters for Digits)

1. Understand the Rules of the Puzzle — Each letter in the puzzle must stand for just one unique digit. The first digit of a number cannot be zero (for example, we cannot write the number forty-five as 045).
2. Analyze the Units Place First — Look at the addition or multiplication in the units column. Find which digits (0-9) satisfy the given condition. For example, if $A + 5 = 2$ (at units place), then $A + 5$ must equal $12$, which gives $A = 7$, carrying over 1 to the tens place.
3. Account for Carry-overs in Tens and Hundreds Places — Apply the carry-over from the units place to the tens column. Write down the equation for the tens column and solve for the remaining unknown letters.
4. Verify Your Digits — Substitute the calculated digits back into the original puzzle. Perform the complete addition or multiplication to ensure all columns align correctly and no rules are violated.

Exam Tip: Why the Divisibility Rules for 3 and 9 Work

Many students memorise that a number is divisible by 9 if the sum of its digits is divisible by 9, but they don't know why. Let's look at a three-digit number $abc$, written as $100a + 10b + c$. We can rewrite this as:

$100a + 10b + c = (99a + a) + (9b + b) + c = (99a + 9b) + (a + b + c) = 9(11a + b) + (a + b + c)$.

Notice that $9(11a + b)$ is always divisible by 9. Therefore, the entire number is divisible by 9 if and only if the remaining part, $(a + b + c)$ (the sum of the digits), is divisible by 9! The same logic applies to the divisibility rule for 3. Remembering this algebraic breakdown will help you tackle complex logic questions in exams without confusion.

Practice Questions with Solutions

• Q: Find the values of letters A and B in the addition: 3A + 25 = B2. A: Step 1: Look at the units column: $A + 5$ ends in 2. Step 2: This means $A + 5 = 12$, which gives $A = 12 - 5 = 7$. Step 3: Carry over 1 to the tens column. Now, look at the tens column: $1 \text{ (carry)} + 3 + 2 = B$. Step 4: This simplifies to $B = 6$. Final answer: A = 7, B = 6.
• Q: Find the values of A and B in the multiplication: AB * 6 = BBB. A: Step 1: The units place multiplication is $B \times 6$, which must end in the digit $B$. The single digits that satisfy this are 0, 2, 4, 6, and 8. Step 2: If $B = 0$, then $AB \times 6 = 000$, which is not a valid 3-digit number because the first digit B cannot be 0. Step 3: Try $B = 4$. The equation becomes $A4 \times 6 = 444$. Step 4: Solve for $A$: $A4 = 444 / 6 = 74$. This matches perfectly, giving $A = 7$ and $B = 4$. Final answer: A = 7, B = 4.
• Q: If the four-digit number 21y5 is a multiple of 9, where y is a digit, find the value of y. A: Step 1: A number is a multiple of 9 if the sum of its digits is divisible by 9. Step 2: Find the sum of the digits of 21y5: $2 + 1 + y + 5 = 8 + y$. Step 3: Since $y$ is a single digit (from 0 to 9), the sum $8 + y$ can only be 9 or 18. Step 4: If $8 + y = 9$, then $y = 1$. If $8 + y = 18$, then $y = 10$, which is not a single digit. Final answer: y = 1.
• Q: Prove that the sum of a three-digit number abc and its cyclic permutations (bca and cab) is always divisible by 111. A: Step 1: Write the three numbers in their generalised forms: $abc = 100a + 10b + c$ $bca = 100b + 10c + a$ $cab = 100c + 10a + b$ Step 2: Add all three expressions together: Sum = $(100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)$ Step 3: Group the like terms: Sum = $(100a + a + 10a) + (10b + 100b + b) + (c + 10c + 100c)$ Sum = $111a + 111b + 111c = 111(a + b + c)$ Step 4: Since 111 is a factor of the sum, the sum is always divisible by 111. Final answer: The sum is represented as 111(a + b + c), proving divisibility by 111.

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