Factorial Symmetry

Number Theory Level 3

$\large a! b! = a! + b!$

Let all the pairs of positive integer solutions of $(a,b)$ satisfying the equation above be $(a_1, b_1) , (a_2, b_2) , \ldots , (a_n , b_n)$ . Find $(a_1 + b_1) + (a_2 + b_2) + \cdots + (a_n + b_n) .$

Notation : $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

The answer is 4.

2 solutions

Timothy Ong
Aug 29, 2016

Without loss of generality, let $a≥b$ . Rearranging: $a!(b!-1)=b!$

Making a! The subject of the formula, we get $a!=b!/(b!-1)$

Now note that since a! Is an integer, $b!-1|b!$

Notice that for any integer $n, n≡1 mod (n-1)$ .

Therefore $b!-1$ is only a factor of $b!$ if it is equals to 1. $∴b!=2,a!=2$

The roots of the equation are hence $a=b=2$ and the sum is 4.

isnt there any general method

abhishek alva - 4 years, 9 months ago

Dino Wibisono
Sep 3, 2016

Note that (a!-1)(b!-1)=a!b!-a!-b!+1 We know that a!b!=a!+b! So (a!-1)(b!-1)=1 Since a! and b! are integers, then the only possible value of (a!-1) and (b!-1) are both 1 which means a!=b!=a=b=2 and the sum is a+b=2+2=4

Factorial Symmetry

The answer is 4.

2 solutions

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