7^6 - 5^6 is divisible by?

solved using number properties.
7^6 - 5^6
is of form \(a^n-b^n.\) this is divisible by a+b only if n is even.
hence this will be divisible by 12. so by 6.
\(7^6 - 5^6= (7^3+5^3)(7^3-5^3)\)
\(a^n-b^n\) is divisible by n for all integers.
So the divisible should have 3. No other option has 3 as factor.
Hence only A

Hey Gladiator59

how are you havent seen you around for quite sometime

In the problem above, where is option E? or GMAT format has changed


Another question: what algebraic formula do we use here ?

thank you and good night!!!

dave13, I'm doing good. Hope you are well yourself.

Looks like the OP forgot to post the option (E).

One very tricky way to solve this question would be - straightaway eliminate option (D) as if it is divisible by 32 then it must be divisible by 16 :-D

Now time to play with the exponent -

7^6 can we written as (7^3)^2 ... And so it boils down to known identity a^2-b^2 = (a+b)*(a-b)

7^6 -5^6
(7^3-5^3) * (7^3+5^3)

You could break your head with trying to remember the identity for the above but better would be to just calculate here as seven cube is 343 and five cube is 125 ....did I ever mention how useful it could be to remember cubes till 10?

From this is becomes clear that 16 and hence 32 do not divide the expression and 6 does.

Good night.