**1. ** On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

7 + 5 = 12 in which ones place is 2.

∴ A = 7

And putting 2 and carry over 1, we get

B = 6

Hence A = 7 and B = 6

**2. **On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

8 + 5 = 13 in which ones place is 3.

∴ A = 5

And putting 3 and carry over 1, we get

B= 4 and C= 1

Hence A = 5, B = 4 and C = 1

**3. **On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

A × A = 6 × 6 = 36 in which ones place is 6.

∴ A = 6

Hence A = 6

**4. **Here, we observe that B = 5 so that 7 + 5 = 12.

Putting 2 at ones place and carry over 1 and A = 2, we get

2 + 3 + 1 = 6

Hence A = 2 and B = 5

**5. **Here on putting B = 0, we get 0 × 3 = 0.

And A = 5, then 5 × 3 = 15

A = 5 and C = 1

Hence A = 5, B = 0 and C = 1

**6. **On putting B = 0, we get 0 × 5 = 0 and A = 5, then 5 × 5 = 25

⇒ A = 5, C = 2

Hence A = 5, B = 0 and C = 2

**7. **Here product of B and 6 must be same as ones place digit as B.

6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24

On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.

For 6 × 7 = 42 + 2 = 44

Hence A = 7 and B = 4

**8. **On putting B = 9, we get 9 + 1 = 10

Putting 0 at ones place and carry over 1, we get

For A = 7 ⇒ 7 + 1 + 1 = 9

Hence A = 7 and B = 9

**9. **On putting B = 7,

7 + 1 = 8

Now A = 4, then 4 + 7 = 11

Putting 1 at tens place and carry over 1, we get

2 + 4 + 1 = 7

Hence A = 4 and B = 7

**10.** Putting A = 8 and B = 1, we get

8 + 1 = 9

Now again we add 2 + 8 = 10

Tens place digit is ‘0’ and carry over 1.

Now 1 + 6 + 1 = 8 = A

Hence A = 8 and B = 1

**11. **Since 21y5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

∴ 2 + 1 + y + 5 = 8 + y

8 + y = 9 ⇒ y = 1

**12. ** Since 31z5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

∴ 3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 9 ⇒ *z *= 0

If 3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 18 ⇒ *z *= 9

Hence 0 and 9 are two possible answers.

**13.** Since 24x is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

∴ 2 + 4 + x = 6 + x

Since x is a digit.

⇒ 6 + x = 6 ⇒ x = 0

⇒ 6 + *x *= 9 ⇒ x = 3

⇒ 6 + x = 12 ⇒ x = 6

⇒ 6 + x = 15 ⇒ x = 9

Thus, *x *can have any of four different values.

**14. ** Since 31z5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of

Since *z *is a digit.

∴ 3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 9 ⇒ *z *= 0

If 3 + 1 + z + 5 = 9 + z

⇒ 9 + *z *= 12 ⇒ *z *= 3

If 3 + 1 + z + 5 = 9 + z

⇒ 9 + *z *= 15 ⇒ *z *= 6

If 3 + 1 + z + 5 = 9 + z

⇒ 9 + *z *= 18 ⇒ *z *= 9

Hence 0, 3, 6 and 9 are four possible answers.