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Difference between revisions of "1986 AHSME Problems/Problem 9"

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\textbf{(E)}\ \frac{7}{10} </math>     
 
\textbf{(E)}\ \frac{7}{10} </math>     
  
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==Solution==
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==Solution 1==
 
Factor each term in the product as a difference of two squares, and group together all the terms that contain a <math>-</math> sign, and all those that contain a <math>+</math> sign. This gives <math>[(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}</math>, which is <math>\boxed{C}</math>.
 
Factor each term in the product as a difference of two squares, and group together all the terms that contain a <math>-</math> sign, and all those that contain a <math>+</math> sign. This gives <math>[(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}</math>, which is <math>\boxed{C}</math>.
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==Solution 2 (Answer Choices)==
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Notice that the numerator of the last factor in this product is <math>99</math>, which is divisible by <math>11</math>.  Also, the denominators of the factors in this product are <math>2^2</math>, <math>3^2</math>, ..., <math>10^2</math>, none of which are divisible by <math>11</math>.  Therefore, the numerator of the answer must be divisible by <math>11</math>, and the only such answer is <math>\boxed{(\mathbf{C}) \frac{11}{20}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 06:02, 5 July 2025

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 $\boxed{(\mathbf{C}) \frac{11}{20}}$.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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