That's something interesting!

$a!b!=a!+b!$ or, $a!b!-a!-b!+1=1$ or, $(a!-1)(b!-1)=1$ .

So, $a!-1=b!-1=1$ or, $a!=b!=2$ or, $a=b=2$

or, $a+b=\boxed{4}$

If we divide through by $\huge b!$ we get,

$\huge a! = \frac {a!}{b!} + 1$

As LHS is integer,

$\huge\frac {a!}{b!}$ must be integer, and it follows that $\huge b! \leq a!$

If, instead, we divide through by $\huge a!$ we get,

$\huge b! = 1 + \frac {b!}{a!}$ , and in the same way we deduce that $\huge a! \leq b!$

Hence $\huge a! = b!$ , giving the unique solution $\huge a! = 2 , a = b = 2$

Alternatively, divide the original equation by $\huge a!b!$ which gives,

$\huge 1 = \frac {1!}{b!} + \frac {1!}{a!}$

Clearly , $\huge a! =b! = 2$

$\huge Thus, a + b = 2 + 2 = \boxed{4} :)$

• a!b! = a! + b! $\Rightarrow$ a!b! - a! = b! $\Rightarrow$ a!(b!-1) = b!
• Then $a! = \frac {b!}{b! - 1}$ : a , b , a! , b! are Positive Integers and $b! \neq 1$
• $\Rightarrow$ Denominator $(b! - 1) = 1$ $\Rightarrow$ b! = 2 $\Rightarrow$ a! = $\frac {2}{2 - 1}$ = 2
• $\Rightarrow$ a = b = 2 $\Rightarrow$ a + b = 2 + 2 = $\boxed{4}$

$a! b! - a! = b!$ so $a! (b! - 1) = b!$

it means that $b! - 1$ divides its consecutive $b!$ . But two consecutive numbers are alwais coprime (its greates common divisor is 1) and this imply that $(b! -1)$ is 1. So $b! = 2$ and the only solution of the latter is $b=2$ .

We can change the role of $a$ and $b$ since the initial equation is symmetric and with the same calculations and resoning we get $a = 2$ .

So $a + b = 4$ .

What is the point of using a & b if the number is the same for each? Seems like a bad question to me.

trial and error...both of the numbers must be small which i found intuitively..... put a =b =1...doesnt work....a=b=2...works!! voila!! not a perfect method to solve though.

I just know the only number were a+b=a×b is 2+2=2×2

The diofantine equation $x+y=xy$ only has two nonnegative solutions, $x=y=0$ and $x=y=2$ , as I and other people show here (I'm the asker).

So $a!=b!=0$ or $a!=b!=2$ . The former is impossible, so $a!=b!=2$ thus $a=b=2$ so $a+b=4$ .

a!b! =a!+b! a!b! - a!-b! =0 a!(b-1)-b!=0 a!b!-b!=b! a!b! = 2b! a! =2 b!=2 a=2 b=2

a!b!=a!+b! => b! = a!/b! +1 => b! = a a-1 a-2...a-b +1 ----(1) any factorial other than 0 and 1 has to be even but if b is either 0 or 1 then there exits no +ve or 0 valued 'a' satisfying (1) or a a-1 a-2 ....a-b has to be odd which is not possible => hence there are 2 cases case 1: a = b+1 b! = a*b!/b! +1 => b! = a +1 => b! = b +1 +1 => b = b! -2 no such b exists. case 2: a = b => b! = 1+1 => b! = 2 => b =2 =a this satisfies all equations => a + b =4

The only possible number that has the same of value of its sum and product is 2.

2!+2!=2!2!

I don't understand your solution