Difference between revisions of "1985 AJHSME Problems/Problem 9"
5849206328x (talk | contribs) (New page: ==Problem== The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math> <math>\text{(A)}\ \frac {1}{10} \...) |
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==Solution== | ==Solution== | ||
| − | {{ | + | If we write out the whole product, then we'll 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> |
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| + | If you don't believe this, then write the product as <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>\frac{1}{10}</math> | ||
==See Also== | ==See Also== | ||
[[1985 AJHSME Problems]] | [[1985 AJHSME Problems]] | ||
Revision as of 21:48, 13 January 2009
Problem
The product of the 9 factors 
Solution
If we write out the whole product, then we'll 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 write the product as ![\[\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}\]](http://latex.artofproblemsolving.com/5/9/9/599ba642a006009d04a0c49f11f7ea8baf8ce4b3.png)
Everything except for the first term is , so the product is
