Suppose m, n are positive integers such that | vedclass

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(c)

We have,

$6^n+2^{n+n} \cdot 3^w+2^n=332$

When $m=4, LHS > RHS$

$	herefore$ Maximum value of $m=3$

When $m=3$,

$6^3+2^3 \cdot

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(c)

We have,

$6^n+2^{n+n} \cdot 3^w+2^n=332$

When $m=4, LHS > RHS$

$	herefore$ Maximum value of $m=3$

When $m=3$,

$6^3+2^3 \cdot 2^n \cdot 3^w+2^n=332$

$2^n\left(8 \cdot 3^w+1ight)=332-216$

$2^n\left(8 \cdot 3^{2 v}+1ight)=116$

$2^n\left(8 	imes 3^w+1ight)=4 	imes 29$

$	herefore n=2$

$3^w+1=29 \Rightarrow 3^w=\frac{7}{2}$

Put $m=2$,

$	herefore 6^2+2^2 \cdot 2^n \cdot 3^{w v}+2^n=332$

$2^n\left(4 \cdot 3^{w v}+1ight)=332-36$

$2^n\left(4 \cdot 3^{w v}+1ight)=296$

$2^n\left(4 \cdot 3^w+1ight)=2^3 	imes 37$

$	herefore 2^n=2^3 	ext { and } 4 \cdot 3^{w v}+1=37$

$n=3 	ext { and } 3^w=9 \Rightarrow w=2$

Hence, $m, n, w$ are positive integer.

$	herefore m^2+m n+n^2=(2)^2+(2)(3)+(3)^2$

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

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