VJesus12 wrote:If a number N is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27?

A. 1
B. 2
C. 3
D. 4
E. 6

The OA is the option C.

Experts, how can I show that C is the correct answer? May you help me here? I'd be thankful.

For N to be divisible by 33 means N/(3*11) = some integer, call it X.

So N = 3*11*X. Raising N to the k means:

N^k = 3^k*11^k*X^k. Dividing this by 27 means dividing by 3^3, so

N^k/(3^3) = (3^k*11^k*X^k)/3^3

Now, there could be other powers of 3 hiding in the X depending on what numbers are actually used, but to guarantee that 3^3 can divide into the numerator, 3^k has to be at least as great as 3^3, meaning that k = 3, C

N is divisible by 33, so N = 33*f (where f is an unknown factor) = 3*11*f
N^k is divisible by 27, so N^k = 27*g (where g is an unknown factor) = 3*3*3*g = (3^3)*g
Combining these 2 facts, we get
[3*11*f]^k = (3^3)*g
3^k * (11*f)^k = (3^3)*g
Comparing similar bases (namely powers of 3 in this case), we see that 3^k = 3^3 (minimum)
Comparing powers we see that k = 3 also
ANSWER = C