Question

Suppose $m,n$ are positive integers such that $6^m+2^{m+n}.3^w+2^n=332$ The value of the expression $m^2+mn+n^2$ is

• A. 7
• B. 13
• C. 19
• D. 21

Answer 1

Answer: (C)

We have,
$6^m+2^{m+n}.3^w+2^n=332$
When $m=4,LHS>RHS$
$\therefore$ Maximum value of $m=3$
When $m=3$,
$6^3+2^3.2^n.3^w+2^n=332$
$2^n(8.3^w+1)=332-216$
$2^n(8.3^w+1)=116$
$2^n(8\times 3^w+1)=4\times 29$
$\therefore n=2$
$8.3^w+1=29$
$\Rightarrow 3^w=\frac{7}{2}$
[not possible $\because w\in I]$
Put $m=2$,
$\therefore 6^2+2^2.2^n.3^w+2^n=332$
$2^n(4.3^w+1)=332-36$
$2^n(4.3^w+1)=296$
$2^n(4.3^w+1)=2^3\times 37$
$\therefore 2^n=2^3$ and $4.3^w+1=37$
$n=3$ and $3^w=9$
$\Rightarrow w=2$
Hence, $m,n,w$ are positive integer
$\therefore m^2+mn+n^2=(2{)}^2+(2)(3)+(3{)}^2$
$=4+6+9=19$