Difference between revisions of "1981 AHSME Problems/Problem 3"

Latest revision as of 23:16, 9 February 2020

What is the least common multiple of ${\frac{1}{x}}$ , $\frac{1}{2x}$ , and $\frac{1}{3x}$ is $\frac{1}{6x}$ ?

The least common multiple of ${\frac{1}{x}}$ , $\frac{1}{2x}$ , 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 $\boxed{\left(D\right) \frac{11}{6x}}$ .

@@ Line 1: / Line 1: @@
+==Problem==
+What is 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>?
 ==Solution==
@@ Line 7: / Line 10: @@
 <math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math>
-The answer is <math>\boxed\left(D\right)</math> <math>\frac{11}{6x}</math>.
+The answer is <math>\boxed{\left(D\right) \frac{11}{6x}}</math>.