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
$\therefore m^{2}+mn+n^{2}=(2)^{2}+(2) (3) +(3)^{2}$
$=4 +6+9 = 19$