Bunuel wrote:

Let's try one more.

If n is an integer >1, is 3^n-2^n divisible by 35?

(1) n is divisible by 15.

(2) n is divisible by 18.

How you like this one?

Explanation to follow.

we need to split 35

35 = 27 + 8

(1) 3^15n-2^15n = (3^3)^5n - (2^3)^5n = 27^5n - 8^5n

considering n to be 1 (odd) we get that 27^5 ends with a 7, and 8^5 ends with an 8. so the last digit will be 9 .

35 cant divide a number ending with 9

consider n to be even now, say 2 i.e. 27^10 - 8^10 .

now we will need some corollaries we get from the binomail theorem. if we have a number as a^n - b^n

then if n is odd , the number is divisible by a - b

if n is even, the number is divisible by a + b

so this makes 27^10-8^10 divisible by 27 + 8 = 35.

thus an even value of n will make the term 3^15n - 2^15n divisible by 35, while an odd value will not be divisble

So 1 is insufficient.

By using the above mentioned rule we can infer that 3^18n-2^18n (27^6n - 8 ^6n) will always be divisible by 35 as the index will always be even.

So 2 is sufficient .

So B.

Exactly. +1 to you.

(1) n is divisible by 15. --> \(3^{15m}-2^{15m}=27^{5m}-8^{5m}\) --> 5m may or my not be even, so insufficient to answer, whether it's divisible by 27+8=35.

(2) n is divisible by 18. --> \(3^{18m}-2^{18m}=27^{6m}-8^{6m}\) --> 6m is even, so 3^18m-2^18m=27^6m-8^6m is divisible by 27+8=35. Sufficient.

B.