If \(n = 2^{10} \cdot 3^{14} \cdot 5^{8}\), how many of the natural-number factors of \(n\) are multiples of 150?

I got 702, but that was wrong...

Guest Apr 6, 2020

#1**0 **

I'm not the best at this, but this is my attempt:

150 = 2 x 3 x 5 x 5 -- so every multiply must have at least 1 two, 1 three, and 2 fives.

So, from the 2^{10} (which consists of 10 twos),

you can select either 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 twos -- 10 different selections.

From 3^{14} (which consists of 14 threes),

you can select either 1, 2, 3, ... 14 threes -- 14 different selections.

From 5^{8}, you must select at least 2 fives, but it could also be 3, 4, 5, 6, 7, or 8 fives

-- 7 different selections.

Multiplying 10 x 14 x 7, I get 980 different ways.

geno3141 Apr 6, 2020