Theorem: the first positive number to have 500 divisors has to be even. How can I get started on this proof? I was thinking originally:
Let $ n $ be odd. (Proving by contradiction) then I dont know.
 A: Let $n$ be the smallest positive number that has $k>1$ divisors and let $n=p_1^{r_1}\times\cdots\times p_s^{r_s}$ be its factorization in primes. If $n$ is odd then $2<p_i$ for $i=1,\dots,s$. Replacing one of the $p_i$ by $2$ results in a smaller number that has the same number of divisors ($k=r_1\times\cdots\times r_s$) so a contradiction is found.
A: To get you started
Assume that the largest number that is divisible by 500 different numbers is $n$, then assume $n$ is not divisible by $2$. and is instead divisible by $x$, which is the smallest positive integer than $n$ can be divided by, hence $x$ must be larger than 2.
To finish the proof (so do not read if you just want to get started)
we find that $\frac{n*2}{x}$ is divisible by 2 and is smaller than $n$, hence $n$ cannot be the smallest number divisible by 500 different numbers
A: If $n$ is odd, let $p$ be its smallest prime divisor, and $p^r$ the greatest power of $p$ that divides $n$. Then, the number
$$\frac{2^rn}{p^r}$$
has the same number of divisors, it is smaller than $n$ and it is even.
A: *

*The smallest number with at least $500$ divisors is $2^6\times 3^2 \times 5^2 \times 7 \times 11 \times 13 = 14414400$

*The smallest number with at exactly $500$ divisors is $2^4\times 3^4 \times 5^4 \times 7 \times 11  = 62370000$

*The smallest number with at exactly $500$ divisors apart from itself is $2^{166}\times 3^2 = 841824943102600080885322463644579019321817144754176$ 
All three of these are even.
A: Hint: Try to construct the smallest number with $k>1$ divisors. If it does not have $2$ as a divisor, can it be the smallest?
