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

Approach: Express the target $2m-3n$ as a linear combination of the two factors, so we maximize it without ever solving for $m,n$ separately.

Step 1: Let $u=m+2n$ and $v=2m+n$, with $uv=27$. Add and subtract: $u+v=3m+3n=3(m+n)$ and $v-u=m-n$. So $m=\dfrac{2v-u}{3}$ and $n=\dfrac{2u-v}{3}$.

Step 2: Then $2m-3n=\dfrac{2(2v-u)-3(2u-v)}{3}=\dfrac{4v-2u-6u+3v}{3}=\dfrac{7v-8u}{3}$. To maximize this with $uv=27$, we want $v$ large and positive and $u$ negative.

Step 3: Integer factor pairs of 27 that also keep $m,n$ integers (need $7v-8u$ divisible by 3): testing $(u,v)=(-9,-3)$ gives $\dfrac{7(-3)-8(-9)}{3}=\dfrac{-21+72}{3}=\dfrac{51}{3}=17$. Testing $(u,v)=(3,9)$ gives $\dfrac{63-24}{3}=13$; $(-3,-9)$ gives $\dfrac{-63+24}{3}=-13$; $(9,3)$ gives $\dfrac{21-72}{3}=-17$.

Step 4: The largest value is $17$, at $(u,v)=(-9,-3)$, i.e. $m=1,\ n=-5$.

Answer: 17.