If $N=2^{5} \times 3^{2} \times 7 \times \square$ and 100 divides evenly into $N$, which of the following numbers could be placed in the box?
(A) 5
(B) 20
(C) 75
(D) 36
(E) 120
Solution 1

For 100 to divide evenly into $N$ and since $100=2^{2} \times 5^{2}$, then $N$ must have at least 2 factors of 2 and at least 2 factors of 5.

From its prime factorization, $N$ already has enough factors of 2 , so the box must provide 2 factors of 5 .

The only one of the five choices which has 2 factors of 5 is 75 , since $75=3 \times 5 \times 5$.

Checking, $2^{5} \times 3^{2} \times 7 \times 75=32 \times 9 \times 7 \times 75=151200$.

Therefore, the only possible correct answer is 75 .

## Solution 2

Multiplying out the part of $N$ we know, we get $N=2^{5} \times 3^{2} \times 7 \times \square=32 \times 9 \times 7 \times \square=2016 \times \square$. We can then try the five possibilities.

$2016 \times 5=10080$ which is not divisible by 100 .

$2016 \times 20=40320$ which is not divisible by 100 .

$2016 \times 75=151200$ which is divisible by 100 .

$2016 \times 36=72576$ which is not divisible by 100 .

$2016 \times 120=241920$ which is not divisible by 100 .

Therefore, the only possibility which works is 75 .

ANswer: (C)