If $m$ and $n$ are integers such that $(m+2n)(2m+n) = 27$, then the maximum possible value of $2m-3n$ is

Given that $m$ and $n$ are integers and $(m+2n)(2m+n) = 27$.

Since both factors are integers, we list all integer factor pairs $(a, b)$ such that $a \times b = 27$:

$(1, 27),\ (27, 1),\ (3, 9),\ (9, 3),\ (-1, -27),\ (-27, -1),\ (-3, -9),\ (-9, -3)$

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Let $m + 2n = a$ and $2m + n = b$.

From these two equations:

$3m = 2b - a \implies m = \dfrac{2b - a}{3}$

$3n = 2a - b \implies n = \dfrac{2a - b}{3}$

For $m$ and $n$ to be integers, both $(2b - a)$ and $(2a - b)$ must be divisible by 3. This happens only when $(a + b)$ is divisible by 3.

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Checking each factor pair for this condition:

$(1, 27) \rightarrow a + b = 28$ -- not divisible by 3

$(27, 1) \rightarrow a + b = 28$ -- not divisible by 3

$(3, 9) \rightarrow a + b = 12$ -- divisible by 3

$(9, 3) \rightarrow a + b = 12$ -- divisible by 3

$(-1, -27) \rightarrow a + b = -28$ -- not divisible by 3

$(-27, -1) \rightarrow a + b = -28$ -- not divisible by 3

$(-3, -9) \rightarrow a + b = -12$ -- divisible by 3

$(-9, -3) \rightarrow a + b = -12$ -- divisible by 3

Only 4 pairs give integer solutions.

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For $(a, b) = (3, 9)$:

$m = \dfrac{18 - 3}{3} = 5$, $\quad n = \dfrac{6 - 9}{3} = -1$

$2m - 3n = 10 + 3 = 13$

For $(a, b) = (9, 3)$:

$m = \dfrac{6 - 9}{3} = -1$, $\quad n = \dfrac{18 - 3}{3} = 5$

$2m - 3n = -2 - 15 = -17$

For $(a, b) = (-3, -9)$:

$m = \dfrac{-18 + 3}{3} = -5$, $\quad n = \dfrac{-6 + 9}{3} = 1$

$2m - 3n = -10 - 3 = -13$

For $(a, b) = (-9, -3)$:

$m = \dfrac{-6 + 9}{3} = 1$, $\quad n = \dfrac{-18 + 3}{3} = -5$

$2m - 3n = 2 + 15 = 17$

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The possible values of $2m - 3n$ are: $-17,\ -13,\ 13,\ 17$

The maximum possible value of $2m - 3n = 17$