Prove that there can't be 985 divisors of $123456...19841985$ 
Numbers from 1 to 1985 are written one after another so they form a new number, $n=123456\ldots19841985$. Prove that there can't be 985 divisors of $n$.

This should be solved on paper, without using programming methods. 
 A: Assuming you mean there can't be exactly $985$ distinct divisors of $n$...
$985=5\times 197$. If $n=p_1^{a_1}\cdots p_r^{a_r}$, with $p_1\lt\cdots\lt p_r$ primes, then the number of divisors of $n$ is $(a_1+1)\cdots(a_r+1)$, so the only way $n$ could have exactly $985$ divisors is if $n=p_1^4p_2^{196}$ for two distinct primes $p_1$ and $p_2$; or $n=p^{984}$ with $p$ a prime.
Since $n$ is divisible by $5$, one of the primes must be $5$; hence, $n$ would be divisible by $25$. However, all multiples of $25$ end in $00$, $25$, $50$, or $75$, which $n$ does not; so $n$ cannot have exactly $985$ distinct divisors.

Robert Israel rightly points out the much simpler argument: $n$ is divisible by $5$ but not $5^2$, so from the formula for number of divisors we immediately see that the number of divisors of $n$ is even, hence not $985$. And even if you don't know the formula for divisors, you can partition the divisors into multiples of $5$ and coprime to $5$, and pair them up to see the number of divisors is even.
