Let $n$ be a positive integer that has exactly three prime divisors, and
at least seven divisors of the form $p^k$, where $p$ is a prime, and $k$ is
a positive integer. Prove that $n$ must be divisible by the cube of an
integer that is larger than $1.$
My work so far :
Let the three prime divisors on $n$ be $a,$ $b,$ $c.$ So, according to the question,
$n = a \times b \times c \times d^p \times e^q \times f^r$ where $a,b,c,d,e,f$ are all prime numbers, and $p,q,r$ are positive powers.
It is also given that $p + q + r = 7$ (as there are 7 divisors)
Moreover, as there are only three prime divisors, them d, e, f are among $a,b,c,$ so,
$n = a \times b \times c\times a^p \times b^q \times c^r$
So, $p+1 + q+1 + r+1 = 7+1+1+1 = 10.$
now, we have 3 positive powers, and and their sum is 10. So, by Pigeonhole principle, there exists at least one power of 4.
It is not quite what question asked for. Is it correct?
Best Answer
$n$ has exactly $3$ prime divisor, call them $p_1$, $p_2$ and $p_3$. Let $n=p_1^ap_2^bp_3^c$. If $a,b,c \leq 2$, then there are at most $6$ divisors of $n$ that are of the form $p^k$, namely $p_1,p_1^2,p_2,p_2^2,p_3,p_3^2$. Since $n$ has seven divisors of the form $p^k$. It must have a divisor of the form $p^3$. Is that clear to you?