Difference between revisions of "1985 AJHSME Problems/Problem 9"
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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> | 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> | ||
| − | Everything except for the first term is <math>1</math>, so the product is <math>\frac{1}{10}</math> | + | Everything except for the first term is <math>1</math>, so the product is <math>\textbf{(A)}\frac{1}{10}</math> |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/ZUZk6J2wCw8 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1985|num-b=8|num-a=10}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | |||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 08:24, 10 January 2023
Contents
Problem
The product of the 9 factors 
Solution
First doing the subtraction, we get ![\[\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}\]](http://latex.artofproblemsolving.com/8/6/0/86094e33694afecfd575933a852fd8d22759e9fe.png)
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 ![\[\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}\]](http://latex.artofproblemsolving.com/5/9/9/599ba642a006009d04a0c49f11f7ea8baf8ce4b3.png)
Everything except for the first term is , so the product is 
Video Solution
~savannahsolver
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. 
