Exercise 15.4

1. Check whether 25110 is divisible by 45.

Solution:

For a number to be divisible by 45, it must be divisible by both and

This is because 45 = 5 × 9, and 5 and 9 are co-primes.

25110 ends in , so it isisn't divisible by 5.

The sum of the digits of 25110 is 2 + 5 + 1 + 1 + 0 = , which isisn't divisible by 9.

Since, 25110 is divisible by both and , it isisn't divisible by 45.

2. Check whether 61479 is divisible by 81.

Solution:

61479 is divisible by 81: YesNo

The sum of the digits of 61479 is 6 + 1 + 4 + 7 + 9 = .

Since 27 isisn't divisible by 9, 61479 isisn't divisible by 9.

However, to be divisible by 81, it must be divisible by 9 twice. 27 isisn't twice divisible by 9, so 61479 isisn't divisible by 81.

3. Check whether 864 is divisible by 36? Verify whether 864 is divisible by all the factors of 36?

Solution:

864 is , so it isisn't divisible by 2.

The sum of its digits is 8 + 6 + 4 = , which isisn't divisible by and . Since it is divisible by 2, 3 and 9, it isisn't divisible by 2 × 2 × 3 × 3 = 36.

Factors of 36 are , , , , , , , , and .

864 isisn't divisible by all its factors: 1, 2, 3, 4, 6, 9, 12, 18, and 36.

4. Check whether 756 is divisible by 42? Verify whether 756 is divisible by all the factors of 42?

Solution:

756 is , so it isisn't divisible by 2. The sum of its digits is 7 + 5 + 6 = , which isisn't divisible by 3 and 9.

Since 756 isisn't divisible by 2 and 3, it isisn't divisible by 6.

Since 42 isisn't a multiple of 6, 756 isisn't divisible by 42.

Factors of 42 are , , , , , , , and .

756 isisn't divisible by all its factors: 1, 2, 3, 6, 7, 14, 21, and 42.

5. Check whether 2156 is divisible by 11 and 7? Verify whether 2156 is divisible by the product of 11 and 7?

Solution:

For divisibility by 11: (2 + 5) - (1 + 6) = - = . So, 2156 isisn't divisible by 11.

For divisibility by 7: 215 - (2×6) = - = .

Further, 20 - (2×3) = - = .

Since 14 isisn't Divisible by 7, 2156 isisn't divisible by 7.

The product of 11 and 7 is and we get: 215677 = .

So, 2156 isisn't divisible by 77.

6. Check whether 1435 is divisible by 5 and 7? Verify if 1435 is divisible by the product of 5 and 7?

Solution:

1435 ends in , so it isisn't divisible by 5.

For divisibility by 7: 143 - (2×5) = 143 - 10 = 133. 13 - (2×3) = 13 - 6 = 7. Since 7 is Divisible by 7, 1435 isisn't divisible by 7.

The product of 5 and 7 is 35. 1435 / 35 = 41. So, 1435 isisn't divisible by 35.

7. Check whether 456 and 618 are divisible by 6? Also check whether 6 divides the sum of 456 and 618?

Solution:

456 is , so it isisn't Divisible by 2. The sum of its digits is 4 + 5 + 6 = , which isisn't divisible by 3. Therefore, 456 isisn't divisible by 6.

618 is , so it isisn't divisible by 2. The sum of its digits is 6 + 1 + 8 = , which isisn't divisible by 3. Therefore, 618 isisn't divisible by 6.

The sum of 456 and 618 is 456 + 618 = .

1074 is , so it isisn't divisible by 2. The sum of its digits is 1 + 0 + 7 + 4 = , which isisn't divisible by 3.

Therefore, 1074 isisn't divisible by 6.

8. Check whether 876 and 345 are divisible by 3? Also check whether 3 divides the difference of 876 and 345?

Solution:

The sum of the digits of 876 is 8 + 7 + 6 = , which isisn't divisible by 3. Therefore, 876 isisn't divisible by 3.

The sum of the digits of 345 is 3 + 4 + 5 = , which isisn't divisible by 3. Therefore, 345 isisn't divisible by 3.

The difference of 876 and 345 is 876 - 345 = .

The sum of the digits of 531 is 5 + 3 + 1 = , which isisn't divisible by 3.

Therefore, 3 divides the difference of 876 and 345.

9. Check whether 22 + 23 + 24 is divisible by 2 or 4 or by both 2 and 4?

Solution:

22 + 23 + 24 = + + = .

28 isisn't divisible by 2 and isisn't divisible by 4.

Therefore, 22 + 23 + 24 is divisible by both 2 and 4.

10. Check whether 322 is divisible by 4 or 8 or by both 4 and 8?

Solution:

322 = 3211 = 911611.

Since 9 leaves a remainder of when divided by 4, 9¹¹ will also leave a remainder of when divided by 4.

Therefore, 322 is notis divisible by 4.

Since it is not divisible by 4, it is notis divisible by 8.

11. If A679B is a 5-digit number divisible by 72, find ‘A’ and ‘B’?

Solution:

If a number is divisible by 72, it must be divisible by both and

This is because 72 = 8 × 9 and 8 and 9 are co-primesits multiples.

For divisibility by 8, the last digits must be divisible by 8. Checking multiples of 8 near 790, we find is divisible by 8. So, B = .

For divisibility by 9, the sum of the digits A + 6 + 7 + 9 + B = A + B + must be divisible by 9.

Since B = 2, we have A + divisible by 9.The next multiple of 9 after 24 is .

So, A + 24 = 27, which means A = .

Therefore, A = 3 and B = 2.