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4-2.Quadratic Equations and Inequations
hard
Suppose $a, b, c$ are positive integers such that $2^a+4^b+8^c=328$. Then, $\frac{a+2 b+3 c}{a b c}$ is equal to
A
$\frac{1}{2}$
B
$\frac{5}{8}$
C
$\frac{17}{24}$
D
$\frac{5}{6}$
(KVPY-2015)
Solution
(c)
We have,
$2^a+4^b+8^c=328$
$\Rightarrow 2^a\left(1+2^{2 b-c}+2^{3 c-a}\right)=2^3 \times 41$
$\therefore \quad a=3$
$\Rightarrow \quad 2^{2 b-3}+2^{3 c-3}=40$
$\Rightarrow \quad 2^{2 b-3}+2^{3 c-3}=2^3 \cdot 5$
$\Rightarrow \quad 3 c-3=3 \Rightarrow c=2, b=4$
$\frac{a+2 b+3 c-3+8+6}{a b c}=\frac{17}{24}$
Standard 11
Mathematics
