Question

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

• A. \(9\)
• B. \(13\)
• C. \(15\)
• D. \(17\)

Hint:

When given a product condition like $(m+2n)(2m+n)=27$ with integer solutions, \begin{itemize} \item Introduce new variables for the linear factors, \item Solve the resulting linear system for $m$ and $n$, \item Impose integrality conditions (often via modular arithmetic), \item Check all factor pairs and then evaluate the required expression. \end{itemize}

Answer 1

The Correct Option is D

We are given \[ (m+2n)(2m+n) = 27. \] Let \[ A = m + 2n,\quad B = 2m + n. \] Then \[ AB = 27. \] Step 1: Express $m$ and $n$ in terms of $A$ and $B$. We have the system \[ \begin{cases} m + 2n = A
[2pt] 2m + n = B \end{cases} \] Multiply the first equation by 2: \[ 2m + 4n = 2A. \] Subtract the second equation: \[ (2m + 4n) - (2m + n) = 2A - B \Rightarrow 3n = 2A - B \Rightarrow n = \dfrac{2A - B}{3}. \] Similarly, multiply the second equation by 2: \[ 4m + 2n = 2B. \] Subtract the first equation: \[ (4m + 2n) - (m + 2n) = 2B - A \Rightarrow 3m = 2B - A \Rightarrow m = \dfrac{2B - A}{3}. \] For $m$ and $n$ to be integers, both $2B - A$ and $2A - B$ must be multiples of 3.
Step 2: Use a modular condition. From $2B - A \equiv 0 \pmod{3}$ and $2A - B \equiv 0 \pmod{3}$: \[ 2B - A \equiv 0 \pmod{3} \quad\Rightarrow\quad 2B \equiv A \pmod{3}, \] \[ 2A - B \equiv 0 \pmod{3} \quad\Rightarrow\quad 2A \equiv B \pmod{3}. \] Add these two congruences: \[ 2B - A + 2A - B \equiv 0 \pmod{3} \Rightarrow A + B \equiv 0 \pmod{3}. \] So $A + B$ must be divisible by 3.
Step 3: List factor pairs of $27$. Since $AB = 27$, possible integer pairs $(A,B)$ are: \[ (1,27),\ (3,9),\ (9,3),\ (27,1), \] and their negative counterparts: \[ (-1,-27),\ (-3,-9),\ (-9,-3),\ (-27,-1). \] Check $A + B \equiv 0 \pmod{3}$: - $(1,27)$: $A+B = 28$ (not divisible by 3) - $(27,1)$: $A+B = 28$ (not divisible by 3) - $(3,9)$: $A+B = 12$ (divisible by 3) - $(9,3)$: $A+B = 12$ (divisible by 3) - $(-3,-9)$: $A+B = -12$ (divisible by 3) - $(-9,-3)$: $A+B = -12$ (divisible by 3) So valid pairs are: \[ (3,9),\ (9,3),\ (-3,-9),\ (-9,-3). \]
Step 4: Compute $(m,n)$ and $2m-3n$ for each valid pair. Using \[ m = \frac{2B - A}{3},\quad n = \frac{2A - B}{3}, \] we evaluate each pair. \underline{Pair $(A,B) = (3,9)$:} \[ m = \frac{2\cdot 9 - 3}{3} = \frac{18 - 3}{3} = 5,\quad n = \frac{2\cdot 3 - 9}{3} = \frac{6 - 9}{3} = -1. \] \[ 2m - 3n = 2\cdot 5 - 3(-1) = 10 + 3 = 13. \] \underline{Pair $(A,B) = (9,3)$:} \[ m = \frac{2\cdot 3 - 9}{3} = \frac{6 - 9}{3} = -1,\quad n = \frac{2\cdot 9 - 3}{3} = \frac{18 - 3}{3} = 5. \] \[ 2m - 3n = 2(-1) - 3\cdot 5 = -2 - 15 = -17. \] \underline{Pair $(A,B) = (-3,-9)$:} \[ m = \frac{2(-9) - (-3)}{3} = \frac{-18 + 3}{3} = -5,\quad n = \frac{2(-3) - (-9)}{3} = \frac{-6 + 9}{3} = 1. \] \[ 2m - 3n = 2(-5) - 3\cdot 1 = -10 - 3 = -13. \] \underline{Pair $(A,B) = (-9,-3)$:} \[ m = \frac{2(-3) - (-9)}{3} = \frac{-6 + 9}{3} = 1,\quad n = \frac{2(-9) - (-3)}{3} = \frac{-18 + 3}{3} = -5. \] \[ 2m - 3n = 2\cdot 1 - 3(-5) = 2 + 15 = 17. \]
Step 5: Choose the maximum value. Possible values of $2m - 3n$ are: \[ 13,\ -17,\ -13,\ 17. \] The maximum among these is \[ \boxed{17}. \]