# 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.

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