How many natural-number factors does $N$ have if $N = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7^2$?
We have a prime factorization of $2^4 \times 3^3 \times 5^2 \times 7^2$
If we add one to each exponent, then add up all the exponents.
$5 + 4 + 3 + 3 = 15$
Therefore there are 15 natural factors.
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Sorry, but that is wrong.
OBJD is close....but....we need to add 1 to each exponent and then take that product....so..
(4+1) (3+1) (2 + 1) ( 2 + 1) =
5 * 4 * 3 * 3 =
180 factors
