# 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> |

+ | |||

+ | 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> | ||

+ | |||

+ | 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 22:48, 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

Everything except for the first term is , so the product is