I posted the picture above to each of our social media outlets, and I've posted my solution below...

The solution to this problem requires a bit of creativity. As you can see above, we have to make a point by "starting over" halfway through. Because

*a*is an integer as defined in the problem, we know that*a!*must be an integer, and then it follows that the other side of the equation must add to an integer. So the fractions must each divide evenly.
As we finish up the problem, we discover that

*a*and*b*are the same variable.
Did you solve the problem another way? I know there is at least one other method to solve (it involves starting by dividing each side by (

*a!b!*). If you have another method, or if you have any questions about my logic, please leave a comment! As always, thanks for playing along.
-Tori

please, could you give me an e-mail to send you my proof ? i'm not really sure if i've done it right

ReplyDeleteSure! You can always reach us at the address: info@centerofmath.org and it will be forwarded to our math intern from there.

DeleteSet x:=a! and y:=b! (x&y are positive integers)

ReplyDeletex+y=xy <=> y(x-1)=x.

So x-1|x but x-1 and x are coprime, and since x is different from zero x-1=1, hence x=2 and y=2/(2-1)=2. Which gives necessarily a=b=2 which are indeed solution.

V. Pantaloni