$n!=n(n-1)(n-2)\cdots 3\times 2\times 1 $ now let

Question

$n!=n(n-1)(n-2)\cdots 3\times 2\times 1 $ now let $5!m!=n! , m\geq 4$ then $n+m=?$

My Try :

case$1$ Let $n=m$: $$5!m!=m! \\ 5! \neq 1$$ so $n\neq m$

case$2$ Let $n>m$ :

$$5!=\dfrac{n!}{m!} \\ 5!=n(n-1)(n-1) . . . (n-m)$$

case$3$ Let $m>n$

$$5!m!=n! \\ m(m-1)(m-2)...(m-n)=\dfrac{1}{5!}=\dfrac{1}{120}$$ since $\dfrac{1}{120}$ is't natural number so $n>m$ Now what ?

Answer 1

This is easy to figure out by brute force. $5!=120$, and the integer $m$ must satisfy the equation $(m+1)(m+2)\ldots(m+r) = 120$. However, for $r \geq 3$ the integer $m$ can be no larger than 5 (because $5^3> 120$) and for $r=2$ the integer $m$ can be no larger than 11. That leaves few cases left to check.

The solution $m=119$ and $n=5!=120$ works for the equation $5!m! = n!$. So $n+m=119+120 = 239$.

As that is the only way to factor 120 into $(m+1)(m+2)\ldots(m+r)$ for some positive $r$ besides 120=5! and $120 = 6\times 5 \times 4$ and that is the only solution for $m \ge 4$.

Answer 2

Hint: Obviously, $n>m$. Now, we have $$ 5! = n(n-1)\cdots(m+1). $$ In how many ways $5!=120$ can be written as a product of consecutive integer(s) starting from $m+1\ge 5$?

Answer 3

You can just write $5!=120=\frac{n!}{m!}=n(n-1)\cdots(m+1).$

Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$

If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10\times11=110<120$ and $11\times12=132>120$).

If $n-m=3$ the solution is given by $120=6\times5\times4$ so that we get $n=6, m=3.$

If $n-m=4$ then $n(n-1)\cdots(n-3)=120$ which implies $n=5,$ and $m=1.$

Same as the last one, if $n-m=5$ then $n(n-1)\cdots(n-4)=120$ which implies $n=5,$ and $m=0.$

The difference can't be bigger because there would be too many factors to obtain $120.$