Stop shouting! pt.2

Wlog $a\ge b\ge 2$ (none of $a,b$ can be $0$ or $1$ ). Then $a=b+k$ with $k\ge 0$

Then $a!=\frac{a!}{b!}+1+\frac{c!}{b!}$ implies $c=b+m\ge 2$ with $m\ge 0$ .

$b!=1+\frac{b!}{(b+k)!}+\frac{(b+m)!}{(b+k)!}$ . Two cases:

If $k> 0$ , then $k>m\ge 0$ , so $a>c\ge b$ .

Then $\frac{a!}{c!}b!=\frac{a!}{c!}+\frac{b!}{c!}+1$ implies $b\ge c$ , so $b=c$ and $a>b=c$ .

Then $a!(b!-1)=2b!$ , so $a!\mid 2b!$ , so $a!\le 2b!\le (b+1)!\implies a\le b+1$ and $b+1\ge a>b=c$ , so $b+1=a$ .

$(b+1)!b!=(b+1)!+2b!\iff (b+1)!=b+3$ , so $b\mid 3\iff b\in\{1,3\}$ , none of which solve the last equation.

If $k=0$ , then $a=b$ . Then $a!^2=2a!+(a+m)!\iff a!=2+\frac{(a+m)!}{a!}$ , so $m!\mid a!-2> 0$ , since $\binom{a+m}{a}=\frac{(a+m)!}{a!m!}$ is a positive integer.

$m!\le a!-2<a!\implies m<a$ . But then $m!\mid a!$ , so $m!\mid 2\iff m\in\{0,1,2\}$ .

If $m=0$ , then $a=b=c\implies a!^2=3a!\iff a!=3$ , impossible.

If $m=2$ , then $a!^2=2a!+(a+2)(a+1)a!\iff a!=a^2+3a+4$ , so $a\mid 4\iff a\in\{1,2,4\}$ . None of these solve the last equation.

If $m=1$ , then $a!^2=2a!+(a+1)a!\iff a!=a+3$ , so $a\mid 3\iff a\in\{1,3\}$ , which gives the only solution $(a,b,c)=(3,3,4)$ .

$Since~ a~ and~ b ~are~ interchangeable,~ we~ can~ safely~ assume~ a=b~ or~ a>b.\\ Note~ that ~n!>0,~ and~so~minimum~c!=1.\\ If~b=1, a!*b!<a!+b! ,~~ \implies~ a=b>1\\ If~a=b=2,~~a!*b!=a!+b!+0,~but~c!\neq 0,~\therefore~a=b=2~not~a~solution.\\ 12=3!*2!=3!+2!+4. ~~But~c! \neq 4. ~~no~solution.\\ Next~3!*3!=36=3!+3!+24~~so~c!=24~~and~c=4.\\ Solution~(a,b,c)=(3,3,4) ~only.\\ a+b+c=3+3+4=\Large~~\color{#D61F06}{10}.$

Let $a, b, c \in \mathbb N \cup \{0\}$ such that

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad a! + b! + c! = a! b!. \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ (\ast)$

Notice that $(\ast)$ can be rewritten as $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad c! + 1 = (a!-1)(b!-1). \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad (\ast \ast)$

For $a \leq 1$ or $b \leq 1$ , $(\ast)$ becomes $1+ c! = 0$ , which is impossible for $c!>0$ .

WLOG, let $b\geq a \geq 2$ .

$\textbf{Case 1}:\ c \leq b$

Dividing $(\ast)$ by $a!$ and $c!$ yields

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \displaystyle{1 + \frac{b!}{a!} + \frac{c!}{a!} = b!}\quad \quad \quad$ and $\quad \quad \quad \displaystyle{1 + \frac{b!}{c!} + \frac{a!}{c!} = \frac{a! b!}{c!}}$

respectively. Both $\displaystyle{\frac{b!}{a!}}$ and $\displaystyle{\frac{b!}{c!}}$ are integers since $b\geq a$ and $b\geq c$ .

Therefore, the former and latter equalities imply $c\geq a$ and $a\geq c$ respectively, i.e. $c = a$ .

Putting $a=c$ into $(\ast\ast)$ yields

$c! + 1 = (c! - 1)(b!-1)\ \Longleftrightarrow\ (c! - 1) + 2 = (c!-1)(b!-1)\ \Longleftrightarrow\ 2 = (c! -1) (b! - 2)$

So $b! - 2$ is either $2$ or $1$ , i.e. either $b! = 4$ or $b! = 3$ , but such $b$ cannot be found.

$\textbf{Case 2}:\ b < c$

Dividing $(\ast)$ by $a!$ and $b!$ yields

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \displaystyle{1 + \frac{b!}{a!} + \frac{c!}{a!} = b!}\quad \quad \quad$ and $\quad \quad \quad \displaystyle{1 + \frac{a!}{b!} + \frac{c!}{b!} = a! }$

respectively. Using the similar argument as Case $1$ , we have $a = b$ . Putting $a = b$ into $(\ast)$ yields

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad c! = b!(b!-2)\quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ (\ast\ast\ast)$

Observe that $\displaystyle{C^c_b = \frac{c!}{b! (c-b)!}}$ is an integer and now we have

$\quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \displaystyle{C^c_b = \frac{c!}{b! (c-b)!} = \frac{b! - 2}{(c-b)!}}\quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ .

Note that $\displaystyle{\frac{b! - 2}{(c-b)!} = C^c_b \geq 1}$ . It follows that $b! \geq (c-b)! + 2 > (c-b)!$ , and so $b \geq c - b$ .

Consequently, $\displaystyle{\frac{b!}{(c-b)!}}$ is an integer. By rewriting this result as

$\quad \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \displaystyle{C^c_b = \frac{b! - 2}{(c-b)!} = \frac{b!}{(c-b)!} - \frac{2}{(c-b)!}}$ ,

we have $(c-b)!$ as a divisor of $2$ . Since $c > b$ , $c - b$ is either $2$ or $1$ .

$\textbf{Sub-case 2 (a)}:$

For $c = b+1$ , putting this result into $(\ast\ast\ast)$ yields $b! - b = 3$ , for which $b$ is a divisor of $3$ . Hence $b$ is either $1$ or $3$ , but only $b = 3$ satisfies this relation.

$\textbf{Sub-case 2 (b)}:$

For $c = b + 2$ , putting this result into $(\ast\ast\ast)$ yields $b! - b^2 - 3b = 4$ .

Clearly, $b$ divides $4$ and so $b = 1, 2 \text{ or } 4$ . However, none of them satisfy $b! - b^2 - 3b = 4$ .

$\textbf{In conclusion}$ , $a = b = 3$ and $c = 4$ and so $a+b+c = 10$ .

for a=0:10

for b=0:10

for c=0:10

if (factorial(a)*factorial(b)==factorial(a)+factorial(b)+factorial(c))

a+b+c

end

end

end

end

Answer: 10