$P$, $Q$, $R$ are three distinct Prime Numbers

$P!$ = $Q$ x $R^Q$ x $P$

Find P, Q, R.


I think it’s

P = 5, Q = 3, and R = 2.

This gives

$5! = 120 = 3 \times 2^3 \times 5$.

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 $R=2$.

  • $\begingroup$ Great answer and explanation! $\endgroup$
    – El-Guest
    Jun 3 '19 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.