Solve Algebraic Equation for Positive Integers

We are given the equation $m + n + mn = 118$, where $m$ and $n$ are positive integers. We need to find the value of $m + n$. This equation can be rewritten by adding 1 to both sides:

$m + n + mn + 1 = 118 + 1$

Now, we can factor the left side:

$m(1+n) + (n+1) = 119$

$(m+1)(n+1) = 119$

Find Factors of 119

Since $m$ and $n$ are positive integers, $m \ge 1$ and $n \ge 1$. Therefore, $m+1 \ge 2$ and $n+1 \ge 2$. We need to find pairs of factors of 119 that are both greater than or equal to 2.

The factors of 119 are 1, 7, 17, and 119. The pairs of factors $(m+1, n+1)$ whose product is 119 are:

• $(7, 17)$
• $(17, 7)$

The pair $(1, 119)$ is excluded because $m+1$ and $n+1$ must be at least 2.

Determine Values of m and n

Case 1: $m+1 = 7$ and $n+1 = 17$ This gives $m = 7 - 1 = 6$ and $n = 17 - 1 = 16$.

Case 2: $m+1 = 17$ and $n+1 = 7$ This gives $m = 17 - 1 = 16$ and $n = 7 - 1 = 6$.

Calculate the Sum m + n

In both cases, the pair of integers is $(6, 16)$ or $(16, 6)$. Let's find the sum $m + n$ for these pairs:

• If $m=6$ and $n=16$, then $m + n = 6 + 16 = 22$.
• If $m=16$ and $n=6$, then $m + n = 16 + 6 = 22$.

Therefore, the value of $m + n$ is uniquely determined to be 22.