Difference between revisions of "1985 AJHSME Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | + | First doing the subtraction, we get <cmath>\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}</cmath> | |
− | + | We notice a lot of terms cancel. In fact, every term in the numerator except for the <math>1</math> and every term in the denominator except for the <math>10</math> will cancel, so the answer is <math>\frac{1}{10}</math>, or <math>\boxed{\text{A}}</math> | |
− | Everything except for the first term is <math>1</math>, so the product is <math>\frac{1}{10}</math> | + | If you don't believe this, then rearrange the factors in the denominator to get <cmath>\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}</cmath> |
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+ | Everything except for the first term is <math>1</math>, so the product is <math>\textbf{(A)}\frac{1}{10}</math> | ||
==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1985|num-b=8|num-a=10}} |
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 17:05, 3 July 2013
Problem
The product of the 9 factors
Solution
First doing the subtraction, we get
We notice a lot of terms cancel. In fact, every term in the numerator except for the and every term in the denominator except for the will cancel, so the answer is , or
If you don't believe this, then rearrange the factors in the denominator to get
Everything except for the first term is , so the product is
See Also
1985 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.