Difference between revisions of "1985 AJHSME Problems/Problem 1"
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Revision as of 17:01, 3 July 2013
Problem
Solution
We could go at it by just multiplying it out, dividing, etc, but there is a much more simple method.
Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by , we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like
Notice that each number is still there, and nothing has been changed - other than the order.
Finally, since each fraction is equal to one, we have , which is equal to
.
Thus, is the answer.
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.