FACTORIALS, FACTORIALS everywhere, not a single SOLUTION to see!!!

It is quite easy actually. Note first that $r<min(k,m,n)$

where $min(x_{1},x_{2},...,x_{n})$ denotes the minimum real out of the $'n'$ real numbers in the brackets.

Without loss of generality, let us suppose that one of the numbers $k$ , $m$ and $n$ is the largest, say $n$ . i.e. $n>m$ and $n>k$ . (Note the strict inequalities) In such a case multiply throughout by $n!$ . We obtain:

$(n)\times(n-1)\times...\times(k+1) + (n)\times(n-1)\times...\times(m+1) + 1$

$= (n)\times(n-1)\times...\times(r+1)$

Remainder left on division by $n$ is $1$ on the left side but $0$ on the right side. This is absurd for equal quantities. So, the assumption was wrong. There is no single maximum $'n'$ .

Next, we suppose that two numbers out of $k,m,n$ are simultaneously maximum say $m$ and $n$ , i.e. $m=n>k$ . (Note the strict inequality).

With the same reasoning, this case fails and so this assumption is wrong. So,the only choice is all the terms are simultaneously equal (and then obviously maximum), i.e. $k=m=n$ .

Now, we get $k!=3\times r!$ (in fact $k!=m!=n!)$

Now if $k-r$ was $2$ or more, atleast one even number must be multiplied to get $k!$ from $r!$ by successive multiplication: in other words $k!$ = $(even)\times r!$ . But that is not so. So, $r$ and $k$ are consecutive, i.e. $k=r+1$ .

Immediately, we get: $r=2$ and $k=3=m=n$ as the only solution.

only solution is (3,3,3,2)