Find the largest integer $n$ such that $n$ has exactly 4 positive divisors and $n$ divides $100!$.

Question

Find the largest integer $n$ such that $n$ has exactly 4 positive divisors and $n$ divides $100!$.

Any hints are greatly appreciated.

Answer 1

A number $n$ with prime decomposition $$n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot \ldots\cdot p_r^{\alpha_r}$$ has $$\sigma_0(n)=(1+\alpha_1)(1+\alpha_2)\ldots(1+\alpha_r)$$ positive divisors. Since we want $\sigma_0(n)=4$ we get the possible exponent spectra $(3)$ and $(1,1)$, i.e., $$n=p^3\quad \vee \quad n=p\,q\ .$$ If $\ p^3\>|\>100!$ then at least three multiples of $p$ (namely $p$, $2p$, $3p$) have to be $\leq100$. The largest $p$ fulfilling this requirement is $31$, hence $n=31^3=29\,791$.

The option $n=p\,q$ lets us choose the largest two primes $\leq100$, which leads to $n=97\cdot 89=8633<29\,791$.