Difference between revisions of "2021 JMPSC Sprint Problems/Problem 1"

Revision as of 08:56, 12 July 2021

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 -

@@ Line 12: / Line 12: @@
 <math>\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6}{18}+\frac{3}{18}+\frac{2}{18}=\frac{11}{18}</math> and <math>3+6+9=18</math>, so the answer is <math>\frac{11}{18}\cdot18=\boxed{11}</math>.
+== Solution 3 ==
+<cmath>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}</cmath><cmath>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}</cmath><cmath>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}</cmath>
+Therefore, the answer is <math>\frac{11}{6}+\frac{22}{6}+\frac{33}{6}=11</math>
+- kante314 -
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