The number `6n^2+6n` for natural number n is always divisible by 6 only (b) 6 and 12 (c) 12 only (d) 18 only

In n is a natural number,n(n + 1)(n + 2) will always divisible by

The product of the predecessor and successor of an odd natural number is always divisible by 2 (b) 4 (c) 6 (d) 8

The product of the predecessor and successor of an odd natural number is always divisible by 2 (b) 4 (c) 6 (d) 8

Prove that 7^(n) - 6n - 1 is always divisible by 36.

A 6-digit number is formed by repeating a 3-digit number: for example, 256256 or 678678 etc. Any number of this form is always exactly divisible by 7 only (b) 11 only (c) 13 only (d) 1001

The expression 2^(6n) - 4^(2n) , where n is a natural number is always divisible by

If n is a natural number,then 9^(2n)-4^(2n) is always divisible by 5 (b) 13 (c) both 5 and 13 (d) none of these

If n be any natural number then by which largest number (n^3-n) is always divisible? 3 (b) 6 (c) 12 (d) 18

If n is a natural number, then 9^(2n)-4^(2n) is always divisible by (a) 5 (b) 13 (c) both 5 and 13 (d) none of these