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 $\frac{17}{24}$

$2^3\times 41=2^a[1+2^{2b-a}+2^{3c-a}]$
$\Rightarrow a=3$ (Since $41$ is a prime number and cannot be split further)
$2^{3c-a}+2^{2b-a}=2^3\times 5$
$3c-a=3$
$2b-3c=2$
$a=3,c=2,b=4$
$\frac{a+2b+3c}{abc}=\frac{17}{24}$.