1985 AJHSME Problems/Problem 9

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Problem

The product of the 9 factors $\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =$

$\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \...$

Solution

First doing the subtraction, we get $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}$

We notice a lot of terms cancel. In fact, every term in the numerator except for the $1$ and every term in the denominator except for the $10$ will cancel, so the answer is $\frac{1}{10}$ , or $\boxed{\text{A}}$

If you don't believe this, then rearrange the factors in the denominator to get $\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}$

Everything except for the first term is $1$ , so the product is $\frac{1}{10}$

1985 AJHSME (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
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1985 AJHSME Problems/Problem 9

From AoPSWiki

Problem

Solution

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