$P$, $Q$, $R$ are three distinct Prime Numbers
$P!$ = $Q$ x $R^Q$ x $P$
Find P, Q, R.
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I think it’s
P = 5, Q = 3, and R = 2.
$5! = 120 = 3 \times 2^3 \times 5$.
We note that
$P! = P \times (P-1)!$, so $(P-1)! = R^Q \times Q$. Noting that one of P, Q, and R had to be at least 5, I noted that $4 = 2^2$ and so $4 \times 2 = 2^3$ was probably a convenient way to include these factors of $P!$.
$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Q\cdot R^Q=3\cdot2^3$, and so $Q=3$ and $R=2$.