Question

A number $x$ factors as $107\cdot109^5$, and $107$ and $109$ are primes. What is the exponent of $109$ in the prime factorization of $x^{11}$?

Answer 1

55

Explanation:

We have x^11 = (107 x 109^5)^11 = 107^11 (109^5)^11 = 107^11 109^55, so our answer is 55.

Answer 2

*55*

Yeah it s very easy Bro!

Explanation:

Watch we have something like 109^5=x

then x^11=109^x

11*5=55

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