If a number (N) is divisible by 33, what will be the minimum value of k, such that Nk should definitely be divisible by 27?
I guess it should be N^k instead of Nk.
Given that N is divisible by 33=3*11. Now, in order N^k to be divisible by 27=3^3 then k must be at least 3 in order N^k to have enough 3's.
Thanks as usual, Bunuel. Could you help me wrap my head around this some more?
So N is divisible by 3*11
and N^k is divisible by 3*3*3.
So why exactly does the minimum value for k only need to be 3? Could you possibly use some example values here?
N^k to be divisible by 3^3 should have 3^3 as a factor and since N has only one 3 in its prime factorization then k must be at least 3.