A 7-digit number ‘538p7q2’ is divisible by 24.

Solution

ATQ,

‘538p7q2’ is divisible by 24.

So, it must be divisible by both ‘3’ and ‘8’.

Divisibility by ‘8’: Last three digits must be divisible by ‘8’.

Divisibility by ‘3’: Sum of the digits of the number must be divisible by ‘3’.

So, 7q2 has to be divisible by 8.

Possible values for q = 1, 5, 9

Again, for the number to be divisible by 3,

(5 + 3 + 8 + p + 7 + q + 2) = (25 + p + q) has to be divisible by 3.

If q = 1, then (26 + p) will be divisible by 3 if p = 1, 4, 7

If q = 5, then (30 + p) will be divisible by 3 if p = 0, 3, 6, 9

If q = 9, then (34 + p) will be divisible by 3 if p = 2, 5, 8

Minimum value of (p × q) from all the combinations = 1 × 1 = 1