# 2021 JMPSC Sprint Problems/Problem 1

## Problem

Compute $\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)(3+6+9)$.

## Solution

Solving the right side gives $3 + 6 + 9 = 18$. Distributing into the left side gives $\frac{18}{3}+\frac{18}{6}+\frac{18}{9}$, so the answer is $6 + 3 + 2 = \boxed{11}$.

## Solution 2 $\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6}{18}+\frac{3}{18}+\frac{2}{18}=\frac{11}{18}$ and $3+6+9=18$, so the answer is $\frac{11}{18}\cdot18=\boxed{11}$.

## Solution 3 $$3 \left(\frac{1}{3}\right) + 3 \left(\frac{1}{6}\right) + 3 \left(\frac{1}{9}\right)=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$$ $$6 \left(\frac{1}{3}\right) + 6 \left(\frac{1}{6}\right) + 6 \left(\frac{1}{9}\right)=2+1+\frac{2}{3}=\frac{22}{6}$$ $$9 \left(\frac{1}{3}\right) + 9 \left(\frac{1}{6}\right) + 9 \left(\frac{1}{9}\right)=3+\frac{3}{2}+1=\frac{33}{6}$$ Therefore, the answer is $\frac{11}{6}+\frac{22}{6}+\frac{33}{6}=11$

- kante314 -

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 