1987 AJHSME Problems/Problem 9: Difference between revisions

Revision as of 20:19, 27 May 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 the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $\boxed{\text{C}}$ .

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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
All AJHSME/AMC 8 Problems and Solutions

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 ==See Also==
-{{AJHSME box|year=1986|num-b=8|num-a=10}}
+{{AJHSME box|year=1987|num-b=8|num-a=10}}
 [[Category:Introductory Number Theory Problems]]