is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Question with Factors

What fraction of the divisors of 9! are odd?

This question was asked in the "Who Wants to be a Mathematician?" and the contestants had 60 seconds to answer. Therefore, we should consider a fast way of solving this. Because we have to do a problem with factors, we have to turn that number into its prime factorization. Here is 9! written out with prime factors: $2^73^45^17^1$.

A factor of 9! is thus written as $2^k3^m5^n7^l$, where all of the constants go from 0 to their corresponding exponents above. So $2^03^25^17^0$ is a factor. Another way to think about it is that it's $2^k \left(3^m5^n7^l\right)$. The power of 2 goes from $2^0 \dots 2^7$, which is 8 options. $$2^0 \cdot \text{factors}\left(3^45^17^1\right)$$ $$2^1 \cdot \text{factors}\left(3^45^17^1\right)$$ $$2^2 \cdot \text{factors}\left(3^45^17^1\right)$$ $$\dots$$ $$2^6 \cdot \text{factors}\left(3^45^17^1\right)$$ $$2^7 \cdot \text{factors}\left(3^45^17^1\right)$$ How many of these are odd? Just one, the one with $2^0 = 1$. Therefore with 8 options and one odd one, $\boxed{\dfrac{1}{8}}$ of all factors are odd.
David Witten

Greatest Common Factor