Exercise 15.2

1. If 345A7 is divisible by 3, supply the missing digit in place of ‘A’.

Solution:

For a number to be divisible by 3, the of its digits must be divisible by 3.

3 + 4 + 5 + A + 7 = + A

The next multiple of 3 after 19 is .

So, 19 + A = 21, which means A =

The next multiple of 3 is which means A = .

The next multiple of 3 is which means A =

Therefore, A can be 2, 5, or 8.

2. If 2791A is divisible by 9, supply the missing digit in place of ‘A’.

Solution:

For a number to be divisible by 9, the sum of its digits must be divisible by .

2 + 7 + 9 + 1 + A = 19 + A

The next multiple of 9 after 19 is which means A =

Therefore, A = 8.

3. Write some numbers which are divisible by 2, 3, 5, 9, and 10 also.

Solution:

A number divisible by 2, 5, and 10 must end in .

For it to be divisible by 3 and 9, the sum of its digits must be divisible by (which also makes it divisible by 3).

Examples: 9095, 180163, 270115, 360380, 450500.

4. 2A8 is a number divisible by 2, what might be the value of A?

Solution:

For a number to be divisible by 2, its last digit must be .

The last digit is , which is .

Therefore, A can be any digit from to .

A = 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.

5. 50B is a number divisible by 5, what might be the value of B?

Solution:

For a number to be divisible by 5, its last digit must be or .

Therefore, B = or .

6. 2P is a number which is divisible by 2 and 3, what is the value of P?

Solution:

For 2P to be divisible by 2, P must be . For 2P to be divisible by 3, 2+P2-P must be divisible by .

If P = 0, 2 + 0 = (not divisible by 3)

If P = 2, 2 + 2 = (not divisible by 3)

If P = 4, 2 + 4 = (divisible by 3)

If P = 6, 2 + 6 = (not divisible by 3)

If P = 8, 2 + 8 = (not divisible by 3)

Therefore, P = .

7. 54Z leaves remainder 2 when divided by 5, and leaves remainder 1 when divided by 3, what is the value of Z?

Solution:

If 54Z leaves a remainder of 2 when divided by 5, Z can be or .

If 54Z leaves a remainder of 1 when divided by 3, 5 + 4 + Z = 9 + Z must leave a remainder of 1 when divided by 3.

If Z = 2, 9 + 2 = , which leaves a remainder of when divided by 3.

If Z = 7, 9 + 7 = , which leaves a remainder of when divided by 3.

Z =

Therefore, Z = 7.

8. 27Q leaves remainder 3 when divided by 5 and leaves remainder 1 when divided by 2, what is the remainder when it is divided by 3?

Solution:

If 27Q leaves a remainder of 3 when divided by 5, Q can be or .

If 27Q leaves a remainder of 1 when divided by 2, Q must be .

Therefore, Q = .

The number is 273 i.e. 2 + 7 + 3 = which is divisible by .

Therefore, 273 is divisible by 3, and the remainder is .