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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?
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
$$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot \underbrace{(2\cdot 4)}_8\cdot \underbrace{(3\cdot 3)}_9\cdot \underbrace{(2\cdot 5)}_{10}=10!$$
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 $10!$.
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.
$$\begin{align} 3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\ &= 6 \cdot 15 \cdot 8! \\ &= 2 \cdot 5 \cdot 9! \\ &= 10 \cdot 9! \\ &= 10! \end{align}$$
It should work if we start factoring consecutive integers out of the expression. You have it boiled down to this $$ 6 \cdot 6^3\cdot5^2 \cdot 4^2 \cdot 7 =$$ $$2\cdot 3 \cdot (6^3 \cdot 5^2 \cdot 4^2 \cdot 7)=2 \cdot 3 \cdot 4\cdot 5\cdot 6\cdot 7\cdot (6^2 \cdot 5\cdot 4)$$ This is a good start, but now we need an 8. we get an 8 by using our 4 and a factor of 2 from one of our 6s. This yields: $$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot(3\cdot6\cdot5)$$ We can get a 9 by using our 3 and a factor of 3 from our remaining 6. What remains:$$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot(2\cdot5)=10!$$ How about that?
Notice, $\color{red}{3}$, $\color{red}{5}$, $\color{red}{7}$ are prime numbers (can't be factorized), then the factorials can be successively reduced as follows $$3!5!7!=(3\cdot 2!)\cdot (5\cdot 4!)\cdot (7\cdot 6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2!)\cdot (4!)\cdot (6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2\cdot 1)\cdot (4\cdot 3!)\cdot (6\cdot 5!)$$ $$=1\cdot 2\cdot \color{red}{3}\cdot4\cdot \color{red}{5}\cdot 6\cdot \color{red}{7}\cdot (3! )\cdot (5!)$$
$$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7\cdot (3\cdot 2\cdot1 )\cdot (5\cdot 4\cdot 3\cdot 2\cdot 1)$$ $$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7\cdot ((2\cdot 4)\cdot(3\cdot 3) \cdot(2\cdot 5) )$$ $$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7 \cdot8\cdot9\cdot 10$$$$=10!=n!$$ $$\color{red}{n=10}$$
One can (and possibly should) be even less clever than Andreas proposes. Just compute 3! 5! 7! (getting 3628800) and then use brute force to find a number whose factorial that is. E.g., obviously the number is bigger than 7!, so start with 8! and work upwards. 8! = 40320, too small. 9! = 362880, too small. 10! = 3628800, just right -- and we're done.
If the number were much larger, requiring a search of a larger range of possible n, you could search in a marginally less-brainless way. E.g., find one n that's too small and one that's too large, then repeatedly try one in the middle and halve the range size.
Note that this requires neither intelligence nor good luck, and doesn't really have anything to do with factorials!
