NCERT Solutions for Class 8 Maths Chapter 14: Playing with Numbers — Free PDF

Chapter 14 is one of the most fun chapters in Class 8 Maths. It explores the general form of numbers, divisibility rules for $2, 3, 5, 9, 10$, and exciting number puzzles where you figure out unknown digits.

Understanding the general form of numbers helps explain why divisibility rules work, turning magic tricks into mathematics. This chapter strengthens logical thinking and number sense.

The chapter has two exercises. Exercise 14.1 covers the general form of numbers, properties related to reversing digits, and divisibility tests with proofs. Exercise 14.2 focuses on letter-for-digit (cryptarithmetic) puzzles — these are logic puzzles where each letter represents a unique digit, and you must figure out which digit each letter stands for. Both exercises require careful, systematic reasoning rather than heavy computation.

General Form of a Number: Any number can be expressed in terms of its digits and their place values.

• A $2$-digit number with tens digit $a$ and units digit $b$: $\overline{ab} = 10a + b$, where $1 \leq a \leq 9$ and $0 \leq b \leq 9$.
  - A $3$-digit number: $\overline{abc} = 100a + 10b + c$, where $1 \leq a \leq 9$.

For example, $357 = 3 \times 100 + 5 \times 10 + 7 = 300 + 50 + 7$.

Why is this useful? The general form lets us prove properties about numbers. For instance, we can prove that the sum of a two-digit number and its reverse is always divisible by $11$ — something that seems magical until you see the algebra behind it.

Reversing Digits — Key Results:
- Sum of a $2$-digit number and its reverse: $\overline{ab} + \overline{ba} = (10a + b) + (10b + a) = 11(a + b)$. This is always divisible by $11$.
- Difference of a $3$-digit number and its reverse: $\overline{abc} - \overline{cba} = (100a + 10b + c) - (100c + 10b + a) = 99(a - c)$. This is always divisible by $99$ (and therefore by $9$ and $11$).

Cryptarithmetic (Letter-for-Digit Puzzles): In these puzzles, each letter represents a distinct digit from $0$ to $9$. The leading digit of any number cannot be $0$. You must find the digit for each letter so that the arithmetic (addition or multiplication) is correct.

**Divisibility by $2$:** A number is divisible by $2$ if its last digit is even ($0, 2, 4, 6, 8$).
Example: $4,738$ is divisible by $2$ (last digit $8$).

**Divisibility by $3$:** A number is divisible by $3$ if the sum of its digits is divisible by $3$.
Example: $5,247$: digit sum $= 5 + 2 + 4 + 7 = 18$. Since $18 \div 3 = 6$, the number is divisible by $3$.

Why does this work? In general form, $\overline{abc} = 100a + 10b + c = 99a + 9b + (a + b + c)$. Since $99a + 9b$ is always divisible by $3$, the number is divisible by $3$ exactly when $a + b + c$ is.

**Divisibility by $5$:** A number is divisible by $5$ if its last digit is $0$ or $5$.
Example: $7,835$ is divisible by $5$ (last digit $5$).

**Divisibility by $9$:** A number is divisible by $9$ if the sum of its digits is divisible by $9$.
Example: $8,613$: digit sum $= 8 + 6 + 1 + 3 = 18$. Since $18 \div 9 = 2$, the number is divisible by $9$.

Why does this work? Same reasoning as divisibility by $3$: $99a + 9b$ is divisible by $9$, so the number is divisible by $9$ exactly when the digit sum is.

**Divisibility by $10$:** A number is divisible by $10$ if its last digit is $0$.
Example: $4,560$ is divisible by $10$.

**Divisibility by $11$:** A number is divisible by $11$ if the difference between the sum of digits in odd positions and the sum of digits in even positions is either $0$ or a multiple of $11$.
Example: $91,828$: Odd positions: $9 + 8 + 8 = 25$. Even positions: $1 + 2 = 3$. Difference $= 22$, which is divisible by $11$.

**Q1. Write $357$ in general form.**

Solution:

In algebraic form: if $a = 3$, $b = 5$, $c = 7$, then $\overline{abc} = 100a + 10b + c$.

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**Q2. Show that the sum of a $2$-digit number and the number obtained by reversing its digits is always divisible by $11$.**

Solution:

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

Since the sum equals $11(a + b)$, it is always divisible by $11$.

For example: $47 + 74 = 121 = 11 \times 11$. Here $a + b = 4 + 7 = 11$, and $11 \times 11 = 121$. Verified.

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**Q3. If $\overline{31z5}$ is divisible by $9$, find the value of $z$.**

Solution:

For divisibility by $9$, the sum of digits must be divisible by $9$:

For $9 + z$ to be divisible by $9$: $z = 0$ (giving sum $= 9$) or $z = 9$ (giving sum $= 18$).

So $z = 0$ or $z = 9$.

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**Q4. Show that the difference of a $3$-digit number and the number formed by reversing its digits is divisible by $99$.**

Solution:

Let the number be $\overline{abc} = 100a + 10b + c$ (where $a > c$ so the difference is positive).

Its reverse: $\overline{cba} = 100c + 10b + a$.

Since the result is $99(a - c)$, it is always divisible by $99$.

Example: $852 - 258 = 594 = 99 \times 6$. Here $a - c = 8 - 2 = 6$. Verified.

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**Q5. If a number $\overline{24x}$ is divisible by both $3$ and $9$, find $x$.**

Solution:

Digit sum $= 2 + 4 + x = 6 + x$.

For divisibility by $9$: $6 + x$ must be divisible by $9$. So $x = 3$ (giving sum $= 9$).

Note: If a number is divisible by $9$, it is automatically divisible by $3$. So $x = 3$ satisfies both conditions.

**Q1. Find the digits $A$ and $B$:**

Solution:

Ones column: $A + B$ gives a units digit of $2$. So either $A + B = 2$ or $A + B = 12$ (with carry $1$).

Tens column: $3 + 2 + \text{carry} = B$ (or $B + 10$ if there is a carry to hundreds).

Case 1: No carry from ones. $A + B = 2$ and $3 + 2 = B$, so $B = 5$. Then $A = 2 - 5 = -3$ — invalid (digits must be $0$-$9$).

Case 2: Carry $1$ from ones. $A + B = 12$ and $3 + 2 + 1 = B$, so $B = 6$. Then $A = 12 - 6 = 6$.

Check: $36 + 26 = 62$. The units digit is $2$ and the tens digit is $6 = B$. Correct.

Answer: $A = 6$, $B = 6$.

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**Q2. Find $A$:**

Solution:

Ones column: $1 + A$ must give units digit $0$. So $1 + A = 10$, giving $A = 9$ with carry $1$.

Tens column: $A + 1 + 1 = B$ (or $B$ with carry). $9 + 1 + 1 = 11$, so we write $1$ and carry $1$, making the result $110$ (a $3$-digit number).

Check: $91 + 19 = 110$. So $A = 9$, $B = 1$, and the sum is $110$.

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**Q3. Find $A$ and $B$:**

Solution:

Using general form: $\overline{AB} + \overline{BA} = (10A + B) + (10B + A) = 11(A + B) = 121$.

So $A + B = 11$.

Also, the sum $121$ has $1$ in the hundreds place, which comes from the carry. Since $A + B = 11$ and both are single digits: possible pairs are $(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)$.

But we also need: ones column gives $1$. $B + A = 11$, so the ones digit is $1$ with carry $1$. Tens column: $A + B + 1 = 12$, so tens digit is $2$ with carry $1$. This works for any pair above.

If we need a unique answer, we check: $\overline{AB}$ is a $2$-digit number so $A \geq 1$. Multiple solutions exist (e.g., $29 + 92 = 121$, $38 + 83 = 121$, etc.).

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**Q4. Check divisibility: Is $7{,}248$ divisible by $3$? By $9$?**

Solution:

Sum of digits $= 7 + 2 + 4 + 8 = 21$.

$21 \div 3 = 7$ (exact). So $7248$ **is divisible by $3$**.
$21 \div 9 = 2$ remainder $3$. So $7248$ **is not divisible by $9$**.

**Example 1. Find the missing digit: $\overline{5x3}$ is divisible by $3$ but not by $9$. Find all possible values of $x$.**

Solution:

Digit sum $= 5 + x + 3 = 8 + x$.

For divisibility by $3$: $8 + x$ must be divisible by $3$.
Possible values: $x = 1$ (sum $= 9$), $x = 4$ (sum $= 12$), $x = 7$ (sum $= 15$).

But NOT divisible by $9$: when $x = 1$, sum $= 9$ (divisible by $9$) — exclude.

So $x = 4$ or $x = 7$.

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**Example 2. Show that the sum of any $3$-digit number $\overline{abc}$ and its two rotations $\overline{bca}$ and $\overline{cab}$ is divisible by $111$.**

Solution:

This is always divisible by $111$.

Example: $123 + 231 + 312 = 666 = 111 \times 6$. Here $a + b + c = 1 + 2 + 3 = 6$. Verified.

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Example 3. Solve the cryptarithmetic puzzle:

Solution:

$\overline{2A} \times A = \overline{9A}$, meaning $(20 + A) \times A = 90 + A$.

$20A + A^2 = 90 + A$
$A^2 + 19A - 90 = 0$

Using trial (since $A$ is a digit $0$-$9$):
- $A = 3$: $9 + 57 - 90 = -24$ (no)
- $A = 4$: $16 + 76 - 90 = 2$ (no)
- Try: $(20 + A) \times A = 90 + A$. Test $A = 5$: $25 \times 5 = 125 \neq 95$. No.

Let me re-check: perhaps the result is a $3$-digit number $\overline{9A\_}$. Re-reading the puzzle — if the result is simply the two-digit number $\overline{9A}$:

$A = 3$: $23 \times 3 = 69 \neq 93$. No.
$A = 7$: $27 \times 7 = 189$ (three digits). No.

Actually, this shows why careful reading matters. If the puzzle intended $\overline{2A} \times A = \overline{9A}$, we need $A$ such that the product is a two-digit number starting with $9$. Testing all single digits shows no solution works for this exact setup — which means the puzzle may have additional constraints. This illustrates that not all puzzles have solutions with the stated constraints.

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**Example 4. A $4$-digit number $\overline{3x4y}$ is divisible by both $2$ and $9$. Find $x$ and $y$.**

Solution:

Divisible by $2$: last digit $y$ must be even. So $y \in \{0, 2, 4, 6, 8\}$.

Divisible by $9$: digit sum $3 + x + 4 + y = 7 + x + y$ must be divisible by $9$.

So $7 + x + y = 9$ or $18$ or $27$.

Since $x, y$ are digits ($0$-$9$): $x + y = 2$ or $x + y = 11$.

Combining with $y$ being even:
- $x + y = 2$: $(x, y) = (0, 2)$ or $(2, 0)$.
- $x + y = 11$: $(x, y) = (3, 8), (5, 6), (7, 4), (9, 2)$.

All six pairs are valid answers.

**Mistake 1: Confusing divisibility by $3$ with divisibility by $9$.**
Every number divisible by $9$ is also divisible by $3$, but NOT vice versa. For example, $12$ (digit sum $3$) is divisible by $3$ but not by $9$.

**Mistake 2: Forgetting that the leading digit cannot be $0$.**
In letter puzzles, if a letter appears as the first digit of a number, it cannot be $0$. For example, in $\overline{AB} + \overline{CD} = \overline{EFG}$, the letters $A$, $C$, and $E$ cannot be $0$.

Mistake 3: Not tracking carries in letter puzzles.
Every column in an addition puzzle can produce a carry of $0$ or $1$ (or rarely $2$ for adding three numbers). Forgetting the carry leads to wrong answers. Always process columns from right to left, recording the carry at each step.

Mistake 4: Assuming digits must be distinct in puzzles.
Unless the problem explicitly states "different digits", two different letters might represent the same digit. Read the problem statement carefully.

Mistake 5: Errors in the general form for large numbers.
For a $4$-digit number $\overline{abcd} = 1000a + 100b + 10c + d$. Students sometimes write $100a$ instead of $1000a$ for the thousands digit.

Mistake 6: Mixing up the reversing-digits results.
The sum of a $2$-digit number and its reverse is divisible by $11$ (not $99$). The difference of a $3$-digit number and its reverse is divisible by $99$ (not $11$ alone). Keep these straight.

Here is a comprehensive table of all divisibility rules you need for this chapter and beyond:

Key relationships:
- Every number divisible by $9$ is also divisible by $3$ (but not vice versa)
- Every number divisible by $10$ is also divisible by both $2$ and $5$
- Divisibility by $6$ requires divisibility by BOTH $2$ and $3$
- Divisibility by $12$ requires divisibility by BOTH $3$ and $4$ (not just $2$ and $6$)

Why digit-sum rules work (proof sketch):
Any number $\overline{abc} = 100a + 10b + c = 99a + 9b + (a + b + c)$. Since $99a + 9b = 9(11a + b)$ is always divisible by $9$ (and $3$), the original number is divisible by $9$ (or $3$) exactly when the digit sum $(a + b + c)$ is.

Q1. Write $6,024$ in general form.

Answer: $6024 = 6 \times 1000 + 0 \times 100 + 2 \times 10 + 4 = 6000 + 0 + 20 + 4$.

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Q2. Find the value of $y$ if $\overline{73y}$ is divisible by $3$.

Answer: Digit sum $= 7 + 3 + y = 10 + y$. For divisibility by $3$: $10 + y$ must be divisible by $3$. So $y = 2, 5,$ or $8$.

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Q3. Show that $\overline{ab} - \overline{ba}$ is divisible by $9$.

Answer: $\overline{ab} - \overline{ba} = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)$. Since the result is $9(a - b)$, it is divisible by $9$.

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Q4. Solve: Find $A$ such that $A + A + A = \overline{BA}$ (where $\overline{BA}$ is a two-digit number).

Answer: $3A = 10B + A$, so $2A = 10B$, giving $A = 5B$. Since $A$ is a single digit: $B = 1, A = 5$ (i.e., $5 + 5 + 5 = 15$). So $A = 5, B = 1$.

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Q5. Is $5,49,162$ divisible by $9$?

Answer: Digit sum $= 5 + 4 + 9 + 1 + 6 + 2 = 27$. Since $27$ is divisible by $9$, yes — $549162$ is divisible by $9$.

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Q6. A number $\overline{x8y4}$ is divisible by both $4$ and $9$. If $x = 2$, find $y$.

Answer: Digit sum $= 2 + 8 + y + 4 = 14 + y$. For divisibility by $9$: $14 + y$ must be divisible by $9$. So $y = 4$ (sum $= 18$). The number is $2844$. Check: $2844 \div 9 = 316$. Correct.

• A $2$-digit number $\overline{ab} = 10a + b$; a $3$-digit number $\overline{abc} = 100a + 10b + c$.
  - Sum of a $2$-digit number and its reverse is always divisible by $11$: $11(a + b)$.
  - Difference of a $3$-digit number and its reverse is always divisible by $99$: $99(a - c)$.
  - Divisibility by $3$ or $9$ depends on the sum of digits; by $2$, $5$, $10$ depends on the last digit.
  - Letter puzzles are solved by working column by column from right to left, tracking carries.
  - Always verify your answer by substituting back into the original problem.
  - This chapter builds number sense and logical reasoning — skills that are valuable well beyond Class 8.

Practise number puzzles on SparkEd for a fun way to sharpen your number sense!