Difference between revisions of "1985 AJHSME Problems/Problem 9"
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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> | 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> | ||
| − | If you don't believe this, then | + | 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>\frac{1}{10}</math> | ||
Revision as of 21:53, 13 January 2009
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
