4
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$Given$:

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

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

Find P, Q, R.

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10
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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 $P!$.

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9
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It's:

$5!=3\cdot2^3\cdot5$

Because:

$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Q\cdot R^Q=3\cdot2^3$, and so $Q=3$ and $R=2$.

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

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