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1987 AJHSME Problems/Problem 9

Revision as of 21:39, 14 February 2009 by 5849206328x (talk | contribs) (→‎Solution)

Problem

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

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

Solution

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.

See Also

1987 AJHSME Problems

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