For how many positive integral values of 'p', the number

Solution

For a number to be divisible by 12, it must be divisible by both 3 and 4.

Divisibility of 3: Sum of each digit of the number must be divisible by 3.

Divisibility of 4: Last two digits of the number must be divisibly by 4.

Last two digits of 78292p = 2p

So, 'p' must be 0, 4 or 8 because only 20, 24 and 28 are divisibly by 4.

Sum of each digit of the number = 7 + 8 + 2 + 9 + 2 + p = 28 + p

When 'p' = 0: sum of digits = 28 + 0 = 28 which is not divisibly by 3. So, 'p' = 0 is not valid.

When 'p' = 4: sum of digits = 28 + 4 = 32 which is not divisibly by 3. So, 'p' = 4 is not valid.

When 'p' = 8: sum of digits = 28 + 8 = 36 which is divisibly by 3. So, 'p'= 8 is valid.

Therefore, for only 1 value of 'p', the given number is divisibly by 12.