1987 AJHSME Problems/Problem 9: Difference between revisions

Revision as of 21:39, 14 February 2009

When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$ , the least common denominator used is

$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$

We want $\text{LCM}(2,3,4,5,6,7)$ , which is $420$ , or choice $\boxed{\text{C}}$ . See here if you don't know how to evaluate LCMs.

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 ==Solution==
-We want <math>\text{LCM}(2,3,4,5,6,7)</math>, which is <math>840</math>, or choice <math>\boxed{\text{C}}</math>.  See [[Least common multiple|here]] if you don't know how to evaluate LCMs.
+We want <math>\text{LCM}(2,3,4,5,6,7)</math>, which is <math>420</math>, or choice <math>\boxed{\text{C}}</math>.  See [[Least common multiple|here]] if you don't know how to evaluate LCMs.
 ==See Also==
 [[1987 AJHSME Problems]]