1981 AHSME Problems/Problem 3: Difference between revisions

Revision as of 15:08, 21 November 2018

The least common multiple of $\displaystyle{\frac{1}{x}}$ , $\frac{1}{2x}$ , and $\frac{1}{3x}$ is $\frac{1}{6x}$ .

$\frac{1}{x}$ = $\frac{6}{6x}$ , $\frac{1}{2x}$ = $\frac{3}{6x}$ , $\frac{1}{3x}$ = $\frac{2}{6x}$ .

$\frac{6}{6x}$ + $\frac{3}{6x}$ + $\frac{2}{6x}$ = $\frac{11}{6x}$

The answer is (D) $\frac{11}{6x}$ .

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 ==Solution==
-The least common multiple of <math>\frac{1}{x}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>.
+The least common multiple of <math>\displaystyle{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>.
 <math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.