1986 AHSME Problems/Problem 9

Problem

The product $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\ldots\left(1-\frac{1}{9^{2}}\right)\left(1-\frac{1}{10^{2}}\right)$ equals

$\textbf{(A)}\ \frac{5}{12}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ \frac{11}{20}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{7}{10}$

Solution 1

Factor each term in the product as a difference of two squares, and group together all the terms that contain a $-$ sign, and all those that contain a $+$ sign. This gives $[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{10})] \cdot [(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4})...(1+\frac{1}{10})] = [\frac{1}{2} \frac{2}{3} \frac{3}{4} ... \frac{9}{10}][\frac{3}{2} \frac{4}{3} \frac{5}{4} ... \frac{11}{10}] = \frac{1}{10} \cdot \frac{11}{2} = \frac{11}{20}$ , which is $\boxed{C}$ .

Solution 2 (Answer Choices)

Notice that the numerator of the last factor in this product is $99$ , which is divisible by $11$ . Also, the denominators of the factors in this product are $2^2$ , $3^2$ , ..., $10^2$ , none of which are divisible by $11$ . Therefore, the numerator of the answer must be divisible by $11$ , and the only such answer is $\boxed{(\mathbf{C})\ \frac{11}{20}}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1986 AHSME Problems/Problem 9

Contents

Problem

Solution 1

Solution 2 (Answer Choices)

See also