Difference between revisions of "1981 AHSME Problems/Problem 3"
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| − | == | + | ==Problem== |
| + | For <math>x\neq0</math>, <math>\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{3x}</math> equals | ||
| − | + | <math> \textbf{(A)}\ \dfrac{1}{2x}\qquad\textbf{(B)}\ \dfrac{1}{6x}\qquad\textbf{(C)}\ \dfrac{5}{6x}\qquad\textbf{(D)}\ \dfrac{11}{6x}\qquad\textbf{(E)}\ \dfrac{1}{6x^3} </math> | |
| − | + | ==Solution== | |
| − | <math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x} | + | The least common multiple of <math>x</math>, <math>2x</math>, and <math>3x</math> is <math>6x</math>, so <math>\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} = \frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x} = \boxed{\left(\mathbf{D}\right) \frac{11}{6x}}</math>. |
| − | + | == See Also == | |
| + | {{AHSME box|year=1981|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:59, 29 June 2025
Problem
For 
, 
equals
Solution
The least common multiple of 
, 
, and 
is 
, so 
+ 
+ 
.
See Also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
