If $m$ and $n$ are two positive integers such that $7m + 11n = 200$, then the minimum possible value of $m + n$ is

To find the minimum value of $m + n$ where $7m + 11n = 200$ and $m, n$ are positive integers:

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Rearranging the equation to isolate $m$:

$7m + 11n = 200$

$7m = 200 - 11n$

$m = \dfrac{200 - 11n}{7}$

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Expressing $m + n$ in terms of $n$:

$m + n = \dfrac{200 - 11n}{7} + n$

$= \dfrac{200 - 11n + 7n}{7}$

$= \dfrac{200 - 4n}{7}$

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Since the coefficient of $n$ is negative in our expression, $m + n$ decreases as $n$ increases. To minimize $m + n$, we need to maximize $n$.

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For $m$ to be a positive integer:

$\dfrac{200 - 11n}{7}$ must be a positive integer

$200 - 11n$ must be divisible by 7

$n < \dfrac{200}{11} \approx 18.18$

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Testing values of $n$:

For $n = 15$:

$200 - 11(15) = 200 - 165 = 35$

$m = \dfrac{35}{7} = 5$ (an integer)

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With $n = 15$ and $m = 5$:

$m + n = 5 + 15 = 20$.