Playing With Numbers — for Class 8

Numbers Are Not Just Values — They Have Structure

Every number you write carries hidden patterns inside it. The digit 3 in 300 doesn’t mean the same thing as the digit 3 in 30 or in 3. That positional meaning — the structure of a number — is what this entire chapter exploits.

Class 8 uses that structure to do something clever: we express numbers in general form, then use algebra to prove divisibility rules you’ve been using since Class 4 without knowing why they work. Once you see why “if the sum of digits is divisible by 9, then the number is divisible by 9,” you’ll never forget it.

This chapter also introduces letters for digits — essentially mini-puzzles where you figure out which digit a letter stands for. These are direct predecessors of the logical reasoning questions in competitive exams. They appear in the Olympiad circuit regularly, and a simplified version showed up in CBSE board papers multiple times in the last five years.

The chapter looks light — no heavy calculations — but the conceptual clarity it demands is surprisingly high. Students who rush through it regret it in Class 9 when number theory questions need this as a base.

Key Terms and Definitions

General Form of a Number — Writing a number in terms of its place values. A two-digit number with tens digit $a$ and units digit $b$ is written as $10a + b$ .

47 in general form: $10 \times 4 + 7$
73 in general form: $10 \times 7 + 3$

A three-digit number with hundreds digit $a$ , tens digit $b$ , units digit $c$ becomes $100a + 10b + c$ .

Reversing a Number — Swapping the positions of digits. The reverse of $10a + b$ is $10b + a$ .

Divisibility — A number $N$ is divisible by $d$ if $N \div d$ leaves remainder 0. We write this as $d \mid N$ .

Cryptarithmetic (Letters for Digits) — Puzzles where each letter represents a unique digit (0–9), and the arithmetic equation holds true. For example, solving $AB + BA = \Box\Box$ where A and B are single digits.

Sum of Digits — For a number like 364, the sum of digits is $3 + 6 + 4 = 13$ . This shows up constantly in divisibility tests for 3 and 9.

Core Concepts with Worked Examples

General Form and Why It Matters

Take any two-digit number, say 36. In general form:

36 = 10 \times 3 + 6

Now reverse it: 63. In general form:

63 = 10 \times 6 + 3

We can subtract these using algebra:

36 - 63 = (10 \times 3 + 6) - (10 \times 6 + 3) = 10(3-6) + (6-3) = -27

The difference is 27 — which is $9 \times 3$ . This tells us something: the difference of a two-digit number and its reverse is always divisible by 9. Let’s prove this formally.

Let the number be $\overline{ab} = 10a + b$ and its reverse be $\overline{ba} = 10b + a$ .

\overline{ab} - \overline{ba} = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

This is always divisible by 9. The quotient is $(a - b)$ , the difference of the digits.

Similarly, their sum:

(10a + b) + (10b + a) = 11a + 11b = 11(a+b)

Always divisible by 11. The quotient is the sum of the digits.

Why Divisibility Tests Work

Test for 9: Take 837. Sum of digits = $8 + 3 + 7 = 18$ , which is divisible by 9. So 837 is divisible by 9.

But why? Write 837 in general form:

837 = 100 \times 8 + 10 \times 3 + 7

= 8(99 + 1) + 3(9 + 1) + 7

= (8 \times 99 + 3 \times 9) + (8 + 3 + 7)

= 9(88 + 3) + 18

Both $9(88+3)$ and $18$ are divisible by 9, so 837 is divisible by 9. Notice: what’s left after factoring out 9 is exactly the sum $8 + 3 + 7$ .

This is the algebraic proof that NCERT expects you to understand — not just memorize.

Divisibility by 11

The test: alternately add and subtract digits from right to left. If the result is 0 or divisible by 11, the number is divisible by 11.

For 2728: Start from right — $8 - 2 + 7 - 2 = 11$ . Divisible by 11. ✓

In general form for a 4-digit number $\overline{abcd}$ :

1000a + 100b + 10c + d = (1001a - b + 11b + 11c - c + d)

Splitting the 1000, 100, 10, 1 coefficients into “multiples of 11” plus “remainders” gives the alternating sum. The proof is a natural extension of the 2-digit case.

Letters for Digits — A Systematic Approach

This is where many students guess randomly. Don’t. Work with constraints.

Example: Solve the addition puzzle where $\overline{AB} + \overline{BA} = \overline{DAD}$ , where A, B, D are distinct digits.

Step 1: Write in general form.

(10A + B) + (10B + A) = 100D + 10A + D

11A + 11B = 101D + 10A

11(A + B) = 101D + 10A

Step 2: Since the left side is divisible by 11, the right side must be too. $101D + 10A \equiv 2D + 10A \pmod{11}$ . (Since $101 = 9 \times 11 + 2$ .)

Step 3: Try small values. $D = 1$ : $11(A + B) = 101 + 10A$ , so $11B = 101 - A$ . Try $A = 1$ : $11B = 100$ , not integer. $A = 2$ : $11B = 99$ , so $B = 9$ . ✓

Answer: $A = 2, B = 9, D = 1$ . Check: $29 + 92 = 121 = \overline{DAD} = 121$ . ✓

Solved Examples: Easy to Hard

Example 1 — Easy (CBSE Class 8)

Q: If $\overline{5x2}$ is divisible by 9, find $x$ .

Sum of digits = $5 + x + 2 = 7 + x$ . For divisibility by 9, we need $7 + x = 9$ or $7 + x = 18$ .

$x = 2$ (since $7 + 2 = 9$ ) or $x = 11$ (not a single digit, rejected).

Answer: $x = 2$

Example 2 — Medium (CBSE Class 8 / Olympiad)

Q: A two-digit number is 4 times the sum of its digits. Find the number.

Let the number be $10a + b$ . Then:

10a + b = 4(a + b)

10a + b = 4a + 4b

6a = 3b

b = 2a

Since $a$ and $b$ are single digits (and $a \neq 0$ ): $a = 1, b = 2$ gives 12; $a = 2, b = 4$ gives 24; $a = 3, b = 6$ gives 36; $a = 4, b = 8$ gives 48.

All four are valid. Verify: $12 = 4(1+2) = 4 \times 3 = 12$ ✓

Example 3 — Hard (Olympiad Level)

Q: Solve: $\overline{AB} \times A = \overline{CAB}$ , where A, B, C are distinct non-zero digits.

$\overline{AB} \times A = \overline{CAB}$ means $(10A + B) \times A = 100C + 10A + B$ .

10A^2 + AB = 100C + 10A + B

10A^2 + AB - 10A - B = 100C

10A(A - 1) + B(A - 1) = 100C

(A - 1)(10A + B) = 100C

(A - 1) \times \overline{AB} = 100C

Since $C$ is a single non-zero digit, $100C$ is between 100 and 900. So $(A-1) \times \overline{AB}$ must give a 3-digit multiple of 100. Try $A = 2$ : $1 \times \overline{2B} = 100C$ , so $\overline{2B} = 100C$ — impossible (two-digit = three-digit).

Try $A = 5$ : $4 \times \overline{5B} = 100C$ . So $\overline{5B}$ must be a multiple of 25. Options: 50 ( $B=0$ , rejected, non-zero) or 75 — but that’s $\overline{7B}$ , not $\overline{5B}$ .

Try $A = 3$ : $2 \times \overline{3B} = 100C$ . $\overline{3B}$ must be a multiple of 50. $\overline{3B} = 30$ gives $B=0$ (rejected). No solution.

Try $A = 4$ : $3 \times \overline{4B} = 100C$ . Options where product is a multiple of 100: $\overline{4B} \in \{$ multiples of 100 divisible by 3 $\}$ … $300/3 = 100$ (three-digit, rejected). No.

Try $A = 9$ : $8 \times \overline{9B} = 100C$ . $\overline{9B}$ must give product as multiple of 100. $\overline{9B} = 25k/2$ … Let $8(90 + B) = 100C$ : $720 + 8B = 100C$ . $8B = 100C - 720$ . $C = 8: 8B = 80, B = 10$ (rejected). $C = 7: 8B = -20$ (rejected). No.

The puzzle has specific constraints — competition problems of this type are designed to have a unique solution, and the approach above (factoring into general form) is the method examiners reward.

CBSE Class 8 NCERT exercises test the straightforward cases — reversals, digit-sum divisibility, simple letter-for-digit. Olympiad papers take the same framework deeper with multi-step constraints. Building your general-form algebra here pays compound interest later.

Exam-Specific Tips

CBSE Class 8

Weightage: This chapter carries roughly 3–5 marks in the annual exam, typically as one 2-mark short answer and one 3-mark application question.

What NCERT actually tests:

Proving divisibility rules using general form (guaranteed 1 question)
Finding missing digits using divisibility conditions
Reversing numbers and their sum/difference properties

Always write the general form explicitly in your solution. Examiners give step marks, and $N = 10a + b$ as the first line guarantees you at least 1 mark even if you make an arithmetic error later.

Olympiad (Math Olympiad / NSO)

Letters-for-digits problems appear regularly. The strategy is always: write everything in general form → collect unknowns on one side → use digit constraints (each digit 0–9, leading digit ≠ 0, all letters distinct) to narrow down.

In competitive exams, when you see a letters-for-digits puzzle with a carry in addition, always check the carry separately for each column from right to left. Carries are either 0 or 1 — that constraint often immediately fixes one of the unknowns.

Common Mistakes to Avoid

Mistake 1: Forgetting leading digit ≠ 0. In $\overline{AB}$ , $A$ cannot be 0 (that would make it a 1-digit number). Students often find $A = 0$ as a mathematical solution and include it — wrong. Always state “since $A$ is the leading digit, $A \neq 0$ .”

Mistake 2: Confusing sum of digits with the number. The divisibility test for 9 uses the sum of digits, not the number itself. $45$ is divisible by 9 because $4 + 5 = 9$ , not because 45 looks like it. Students get confused when the sum of digits is itself a two-digit number — just keep summing. $999$ : $9+9+9 = 27$ , $2+7 = 9$ . Yes, divisible by 9.

Mistake 3: Claiming uniqueness without checking all cases. When asked “find the digit,” check every possible value, not just the first one that works. The question “find $x$ if $\overline{2x7}$ is divisible by 3” has $x = 2$ (sum = 11, not divisible), $x = 5$ (sum = 14, not divisible), $x = 8$ (sum = 17, not divisible)… Wait — $2+x+7 = 9+x$ divisible by 3 means $x = 0, 3, 6, 9$ . Multiple answers possible!

Mistake 4: Wrong application of the 11 test. The alternating sum for 11 goes from right to left: units gets $+$ , tens gets $-$ , hundreds gets $+$ , and so on. Many students go left to right and sometimes get the same answer (when the number of digits is even) and sometimes a wrong answer (odd digits). Fix: always start from the rightmost digit.

Mistake 5: Not verifying the answer. Letters-for-digits problems must be verified by substituting back into the original puzzle. You might satisfy the algebraic equation but violate a constraint (like two letters getting the same digit). Always plug back in.

Practice Questions

Q1. Write 786 in general form and find which of the digits (7, 8, 6) contributes how much to the value.

$786 = 100 \times 7 + 10 \times 8 + 6 = 700 + 80 + 6$

Digit 7 contributes 700, digit 8 contributes 80, digit 6 contributes 6.

Q2. Check whether 5765 is divisible by 11.

Alternating sum (right to left): $5 - 6 + 7 - 5 = 1$

Since 1 is not divisible by 11, 5765 is not divisible by 11.

Q3. The sum of a two-digit number and its reverse is 121. Find all possible numbers.

Let the number be $10a + b$ . Sum with reverse: $(10a+b) + (10b+a) = 11(a+b) = 121$ .

So $a + b = 11$ . Possible pairs $(a,b)$ where both are single digits and $a \neq 0$ : $(2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)$ .

Numbers: 29, 38, 47, 56, 65, 74, 83, 92. All eight are valid answers.

Q4. If $\overline{31a}$ is divisible by both 3 and 9, find $a$ .

For divisibility by 9 (which also ensures divisibility by 3): $3 + 1 + a = 4 + a$ must be divisible by 9.

$4 + a = 9 \Rightarrow a = 5$ . (Next would be $4 + a = 18 \Rightarrow a = 14$ , not a digit.)

$a = 5$ , and the number is 315.

Q5. Prove that the difference of a three-digit number $\overline{abc}$ and its reverse $\overline{cba}$ is divisible by 99. What is the quotient?

$\overline{abc} = 100a + 10b + c$

$\overline{cba} = 100c + 10b + a$

Difference: $(100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c)$

This is always divisible by 99. The quotient is $(a - c)$ , the difference of the hundreds and units digits.

Q6. Solve: $\overline{AB} + \overline{BA} = \overline{BB}$ where A and B are distinct non-zero digits.

$(10A + B) + (10B + A) = 11B + 11A = 11(A + B)$

$\overline{BB} = 10B + B = 11B$

So $11(A + B) = 11B \Rightarrow A + B = B \Rightarrow A = 0$ .

But $A$ is a non-zero digit (leading digit of $\overline{AB}$ ). No solution exists under the given constraints.

Q7. Find digits $P$ and $Q$ such that: $\overline{PQ} \times 6 = \overline{QQQ}$ .

$\overline{QQQ} = 111Q = 3 \times 37 \times Q$

$(10P + Q) \times 6 = 111Q$

$60P + 6Q = 111Q$

$60P = 105Q$

$4P = 7Q$

Since $P$ and $Q$ are single digits: $Q = 4, P = 7$ (i.e., $4P = 28 = 7Q$ ).

Check: $74 \times 6 = 444 = \overline{QQQ}$ ✓

$P = 7, Q = 4$

Q8. A 3-digit number is divisible by 4. Its hundreds digit is 3 times its units digit, and its tens digit is 2. Find all such numbers.

Let units digit = $u$ . Hundreds digit = $3u$ . Tens digit = 2. Number = $\overline{3u\,2\,u}$ .

For $3u$ to be a single digit: $u \in \{1, 2, 3\}$ .

Numbers: $u=1$ : 321; $u=2$ : 622; $u=3$ : 923.

Divisibility by 4 — check last two digits: 21 ÷ 4 = no; 22 ÷ 4 = no (22 = 4×5 + 2); 23 ÷ 4 = no.

Hmm — let me rewrite: $u=1$ : last two digits are 21, not divisible by 4. $u=2$ : last two digits are 22, not divisible by 4. $u=3$ : last two digits are 23, not divisible by 4.

No solution exists. This teaches an important lesson — not every setup has an answer, and correctly concluding “no solution” with a clean proof is a full-marks answer.

FAQs

Why do we write numbers in general form — can’t we just use the number directly?

The number itself is a specific value. General form lets us do algebra on digits. When you write $10a + b$ , you can manipulate $a$ and $b$ independently — which is how we prove divisibility rules work for all numbers, not just one specific example.

What is the difference between divisibility and division?

Division always gives a result (quotient + remainder). Divisibility is just the question: is the remainder zero? So 17 ÷ 5 = 3 remainder 2, meaning 17 is not divisible by 5.

Does the divisibility test for 9 work if the sum of digits is itself a large number?

Yes — keep applying the sum-of-digits rule recursively. If the digit sum is 27, then $2 + 7 = 9$ , divisible by 9. This recursive property works because the rule itself is based on the same algebraic structure at every level.

In letters-for-digits, can two different letters have the same digit?

By convention in NCERT problems, different letters represent different digits. The problem statement will say “where A, B, C are different digits.” Always read this constraint carefully before solving.

How is playing with numbers related to Chapter 7 (Cubes and Cube Roots) or other chapters?

The general-form algebra you build here is directly used when proving properties of perfect squares and cubes. For example, proving that a perfect square cannot end in 2, 3, 7, or 8 uses the same digit-level reasoning.

Will divisibility rules appear in JEE?

Not as direct questions at JEE Main level, but number theory in JEE Advanced often involves modular arithmetic, which is precisely “what is the remainder when $N$ is divided by $k$ .” The divisibility rules are a conceptual stepping stone to that.

Can a number be divisible by both 2 and 3 but not by 6?

No. If a number is divisible by both 2 and 3, it is necessarily divisible by $2 \times 3 = 6$ (since 2 and 3 are coprime). This is a useful fact for solving problems quickly.

What if the alternating sum test for 11 gives a negative result?

That’s fine. If the alternating sum is $-11$ or $-22$ etc., the number is still divisible by 11. Divisibility cares about the magnitude, not the sign. For example, $506$ : $6 - 0 + 5 = 11$ . $605$ : $5 - 0 + 6 = 11$ . Both divisible by 11.