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(C) Given the equation $2^a + 2^{2b} + 2^{3c} = 328$. Since $328 = 256 + 64 + 8 = 2^8 + 2^6 + 2^3$,we test powers of $2$. We observe that $328 = 2^3 +

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$\frac{17}{24}$

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(C) Given the equation $2^a + 2^{2b} + 2^{3c} = 328$. Since $328 = 256 + 64 + 8 = 2^8 + 2^6 + 2^3$,we test powers of $2$. We observe that $328 = 2^3 + 2^6 + 2^8$. Comparing $2^a + 2^{2b} + 2^{3c} = 2^3 + 2^6 + 2^8$,we can set $a=3$,$2b=6 \Rightarrow b=3$,and $3c=8$ (which is not an integer). Alternatively,let us re-evaluate: $2^a + 2^{2b} + 2^{3c} = 328$. If $c=1$,$2^a + 2^{2b} = 328 - 8 = 320$. $2^a(1 + 2^{2b-a}) = 320 = 2^6 	imes 5$. So $a=6$,$1 + 2^{2b-6} = 5$ $\Rightarrow 2^{2b-6} = 4 = 2^2$ $\Rightarrow 2b-6=2$ $\Rightarrow b=4$. Thus,$(a, b, c) = (6, 4, 1)$. Check: $2^6 + 4^4 + 8^1 = 64 + 256 + 8 = 328$. This works. Now calculate $\frac{a + 2b + 3c}{abc} = \frac{6 + 2(4) + 3(1)}{6 	imes 4 	imes 1} = \frac{6 + 8 + 3}{24} = \frac{17}{24}$.

Suppose $a, b, c$ are positive integers such that $2^a + 4^b + 8^c = 328$. Then,$\frac{a + 2b + 3c}{abc}$ is equal to

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