Let m and n be two positive integers such that m + n + mn = 118. Then the value of m + n is

Question

Prepp Team · Accepted Answer

Integers Equation: Solving m + n + mn = 118

We are given an equation involving two positive integers, $m$ and $n$: $m + n + mn = 118$. Our goal is to find the value of $m + n$.

The given equation is:

$m + n + mn = 118$

This type of equation, involving the sum and product of two variables, can often be solved by adding a constant to both sides to factor it into the form $(m+a)(n+b) = k$. In this case, we can add 1 to both sides:

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

Rearranging the terms on the left side, we get:

$mn + m + n + 1 = 119$

The left side can be factored by grouping:

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

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

Now, we need to find pairs of factors of 119. Since $m$ and $n$ are positive integers, we know that $m \ge 1$ and $n \ge 1$. This implies that $m + 1 \ge 2$ and $n + 1 \ge 2$.

Let's find the factors of 119. We can check for small prime divisors:

119 is not divisible by 2 (it's odd).
The sum of digits is $1+1+9=11$, which is not divisible by 3, so 119 is not divisible by 3.
119 does not end in 0 or 5, so it's not divisible by 5.
Let's check 7: $119 \div 7$. We know $7 \times 10 = 70$ and $119 - 70 = 49$. Since $7 \times 7 = 49$, $119 = 70 + 49 = 7 \times 10 + 7 \times 7 = 7 \times (10+7) = 7 \times 17$.

So, the factors of 119 are 1, 7, 17, and 119.

We are looking for pairs of factors $(m+1, n+1)$ such that their product is 119, and both factors are greater than or equal to 2. The possible pairs are:

$m+1$	$n+1$	$m = (m+1) - 1$	$n = (n+1) - 1$	Are $m, n$ positive integers?	$m+n$
7	17	$7 - 1 = 6$	$17 - 1 = 16$	Yes (6 > 0, 16 > 0)	$6 + 16 = 22$
17	7	$17 - 1 = 16$	$7 - 1 = 6$	Yes (16 > 0, 6 > 0)	$16 + 6 = 22$

In the first case, $m=6$ and $n=16$. Both are positive integers. Let's check the original equation: $6 + 16 + 6 \times 16 = 22 + 96 = 118$. This is correct.

In the second case, $m=16$ and $n=6$. Both are positive integers. Let's check the original equation: $16 + 6 + 16 \times 6 = 22 + 96 = 118$. This is correct.

In both valid cases where $m$ and $n$ are positive integers, the value of $m+n$ is 22.

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