# Find the Factorial From the Given Prime Relationship

$$Given$$:

$$P$$, $$Q$$, $$R$$ 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 $$P!$$.

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 $$R=2$$.

• Great answer and explanation! Jun 3 '19 at 15:39