# Problem of the Week Problem E and Solution Bilalâ€™s Choices

## Problem

Bilal chooses two distinct positive integers. He adds the product of the integers to the sum of the integers, and then adds $$1$$. He finds that the result is equal to $$196$$.

Determine all possible pairs of integers that Bilal could have chosen.

## Solution

Let $$x$$ and $$y$$ represent the two positive integers that Bilal chooses. Since the integers are distinct, $$x\not =y$$. Let $$x<y$$. That is, let $$x$$ represent the smaller of the two integers.
The product of the two integers is $$xy$$ and the sum is $$(x+y)$$.
Bilal adds the product of the numbers to the sum of the numbers and then adds $$1$$, and the result is $$196$$. Thus, $xy + x+ y + 1 = 196$ Factoring the left side, by grouping the first two terms and the last two terms, we get \begin{aligned} x(y+1)+1(y+1)&=196\\ (x+1)(y+1)&=196 \end{aligned}

Since $$x$$ and $$y$$ are positive integers, then $$x+1$$ and $$y+1$$ are positive integers. Thus, we are looking for a pair of positive integers whose product is $$196$$. There are four ways to factor $$196$$ as a product of two positive integers: $196=1\times 196=2\times 98=4\times 49=7\times 28=14\times 14$

For the product $$196=1\times 196$$, we have $$x+1 = 1$$ and $$y+1 = 196$$. Thus, $$x=0$$ and $$y=195$$. Since the required numbers are positive integers, this solution is inadmissible.

For the product $$196=2\times 98$$, we have $$x+1 = 2$$ and $$y+1 = 98$$. Thus, $$x=1$$ and $$y=97$$. This is a valid solution.

For the product $$196=4\times 49$$, we have $$x+1 = 4$$ and $$y+1 = 49$$. Thus, $$x=3$$ and $$y=48$$. This is a valid solution.

For the product $$196=7\times 28$$, we have $$x+1 = 7$$ and $$y+1 = 28$$. Thus, $$x=6$$ and $$y=27$$. This is a valid solution.

For the product $$196=14\times 14$$, we have $$x+1 = 14$$ and $$y+1 = 14$$. Thus, $$x=13$$ and $$y=13$$. Since the required numbers are distinct, this solution is inadmissible.

Therefore, there are three pairs of distinct positive integers that Bilal could have chosen: $$1$$ and $$97$$, $$3$$ and $$48$$, or $$6$$ and $$27$$. It can be shown that these three pairs do indeed each give the required result.