Find $n$, where its factorial is a product of factorials

I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.

I have tried simplifying as follows:

$$\begin{array}{} 3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\ (3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\ 6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\ 6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\ \end{array}$$

I really didn't see this helping me.

I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.

Any ideas?

Solution 1:

We have

$$3!\cdot 5!\cdot 7!=(1\cdot 2\cdot 3)\cdot (1 \cdot 2\cdot 3\cdot 4\cdot 5)\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7,$$

and combining some of those gives

$$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot \underbrace{(2\cdot 4)}_8\cdot \underbrace{(3\cdot 3)}_9\cdot \underbrace{(2\cdot 5)}_{10}=10!$$

Solution 2:

If the formula is true it can only possibly be $8!$, $9!$, or $10!$ because $11!$ and larger have a factor of $11$. $8!$ doesn't have a large enough power of $3$, and $9!$ doesn't have a large enough power of $5$, so if the formula holds it must be $10!$.

Solution 3:

You already have some clever answers; here's less clever approach. You want $$ 3!\cdot5!\cdot7!=n!=7!\cdot8\cdot9\cdots\cdot n, $$ so, cancelling 7! and computing that $3!\cdot 5!=6\cdot120=720$, you want $$ 720=8\cdot9\cdots\cdot n. $$ Now start dividing both sides by 8, then 9, etc. until you get the answer. Dividing by 8 gives you $90=9\cdots n$, and then you either see the answer already or divide both sides by 9 to get the result.