Algebra madness

It was given that -
$x+xy=391$
$\therefore x(y+1)=391$

Now, the question asks for integeral values. Therefore, we find out the factors of 391, which are $1,17,23,391$

So, $x$ can be 391. Then $y$ will be zero. But the question asks for +ve integers. Therefore, the only possibility is 17 and 23.

$391 = 23\times 17= x(y+1)$

$\implies x=23\text{ and }y=16$

$x+y=23+16=\boxed{39}$

This is a relatively hard problem(computer only allowed level 1s so yeah) that involved a lot of insight. Many attempted the method of trial and error which was very time consuming. The preferable method is to factor out the expression. x+xy=391 turns into x(1+y)=391 Although simpler, it still is pretty far from the resolution. The hardest part about this problem is the number 391 and how it works. I had to think it over for about 30-45 minutes before getting this. The final solution is that 391 is equal to 400-9 right? We then use the formula a^2 -b^2 = (a+b)(a-b) So 400-9 equals to 20^2 - 3^2 and based on the above formula provided. We can conclude that 20^2 - 3^2= (20+3)(20-3) which is equal to 391.

Then we insert the above expression and replace 391 with it so x(y+1)=(20+3)(20-3). x(y+1)=23 times 17 Then we get x=23 and (y+1)=17 and it just so happens that x>y in this process. The final step is that we know that 23 is x and y is 16. Meaning that x+y= 23+16

Therefore: the answer to x+y when x is greater than y is "39"