Answer to Question #150273 in Combinatorics | Number Theory for Khan Asf

Question

Given non-negative integers such that (108^a) ⋅ (288^b) ⋅ (36^c) divided by 6^220. Find the smallest possible value of a + b + c.

Answer 1

$108=2^2\cdot 3^3$

$288=2^5\cdot 3^2$

$36=2^2\cdot 3^2$

So $108^a\cdot 288^b\cdot 36^c=2^{2a+5b+2c}\cdot 3^{3a+2b+2c}$

$6^{220}=2^{220}\cdot 3^{220}$

So $2a+5b+2c\ge 220$ and $3a+2b+2c\ge 220$

Denote $a+b+c$ by $u$, then we have $2u+3b\ge 220$ and $2u+a\ge 220$. Since $c\ge 0$, we also have $u\ge a+b$.

$u=80,a=60,b=20$ (then $c=0$) satisfied these 3 inequalities.

If we take $u<80$, then $b\ge\frac{220-2u}{3}>20$ and $a\ge 220-2u>60$, and we obtain $a+b>60+20=80>u$. It is impossible, so smallest value of $u$ is $80$, when $a=60, b=20, c=0$

Answer: $80$