Difference between revisions of "1987 AJHSME Problems"
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== Problem 9 == | == Problem 9 == | ||
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+ | When finding the sum <math>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}</math>, the least common denominator used is | ||
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+ | <math>\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040</math> | ||
[[1987 AJHSME Problems/Problem 9|Solution]] | [[1987 AJHSME Problems/Problem 9|Solution]] |
Revision as of 21:40, 14 February 2009
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 See also
Problem 1
Problem 2
Problem 3
Problem 4
Martians measure angles in clerts. There are clerts in a full circle. How many clerts are there in a right angle?
Problem 5
The area of the rectangular region is
Problem 6
The smallest product one could obtain by multiplying two numbers in the set is
Problem 7
Problem 8
Problem 9
When finding the sum , the least common denominator used is