If n even, `(6^n-1)` is divisible by 6 (b) 30 (c) 35 (d) 37

AA n in N, 49^n+16n-1 is divisible by (A) 64 (B) 49 (C) 132 (D) 32

7^(6n)-6^(6n), where n is an integer >0, is divisible by 13 (b) 127 (c) 559 (d) All of these

AA n in N,49^(n)+16n-1 is divisible by (A)64 (B) 49(C)132 (D) 32

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

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

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

A) 49^(n) + 16n +1 is divisible by A ( n in N) B) 3^(2n) + 7 is divisible by B ( n in N) C) 4^(n ) - 3n -1 is divisible by C ( n in n) D) 3^( 3n ) - 26 n - 1 is divisible by D (n in N) then the increasing order of A, B, C, D is

If n is any natural number,then 6^(n)-5^(n) always ends with (a) 1 (b) 3 (c) 5 (d) 7 [Hint: For any n in N,6^(n) and 5^(n) end with 6 and 5 respectively.Therefore,6^(n)-5^(n) always ends with 6-5=1.