Difference between revisions of "1985 AJHSME Problem 1"
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The answer is <math>\text{(A) 1}.</math> | The answer is <math>\text{(A) 1}.</math> | ||
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+ | ==Solution== | ||
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+ | the numeretor is 3*5*7*9*11, so is the denominator so (3*5*7*9*11)/(3*5*7*9*11) | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:57, 19 February 2023
Problem
Solution
The answer is
Solution
the numeretor is 3*5*7*9*11, so is the denominator so (3*5*7*9*11)/(3*5*7*9*11)
Video Solution
~savannahsolver
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 0 |
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.