Question

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

• A. $\frac{1}{2}$
• B. $\frac{5}{8}$
• C. $\frac{17}{24}$
• D. $\frac{5}{6}$

Answer 1

The correct option is C $\frac{17}{24}$

$2^a+4^b+8^c=328\Rightarrow 2^a(1+2^{2b-a}+2^{3c-a})=2^3.41$
compare $a=3$
$2^{2b-a}+2^{3c-a}=2^3+2^5$
Either $2b-a=3$ and $3c-a=5\Rightarrow c=\frac{8}{3}$
This is not possible.
Or $2b-a=5$ and $3c-a=3$
$\Rightarrow b=4,c=2$

$\therefore a=3,b=4,c=2$
$\Rightarrow \frac{a+2b+3c}{abc}=\frac{17}{24}$