1985 AJHSME Problems/Problem 9

Revision as of 22:53, 13 January 2009 by 5849206328x (talk | contribs) (Solution)


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

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


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 $\frac{1}{10}$

See Also

1985 AJHSME Problems

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