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imc

prealgebra junior

Problem

What is the value of

\frac{( 1 + \frac{1}{3} ) ( 1 + \frac{1}{5} ) ( 1 + \frac{1}{7} )}{( 1 - \frac{1}{3 ^{2}} ) ( 1 - \frac{1}{5 ^{2}} ) ( 1 - \frac{1}{7 ^{2}} )} ?

(A)

3

(B)

2

(C)

15

(D)

4

Solution

We apply the difference of squares to the denominator, and then regroup factors:

\frac{( 1 + \frac{1}{3} ) ( 1 + \frac{1}{5} ) ( 1 + \frac{1}{7} )}{( 1 - \frac{1}{3 ^{2}} ) ( 1 - \frac{1}{5 ^{2}} ) ( 1 - \frac{1}{7 ^{2}} )} = \frac{( 1 + \frac{1}{3} ) ( 1 + \frac{1}{5} ) ( 1 + \frac{1}{7} )}{( 1 + \frac{1}{3} ) ( 1 + \frac{1}{5} ) ( 1 + \frac{1}{7} ) \cdot ( 1 - \frac{1}{3} ) ( 1 - \frac{1}{5} ) ( 1 - \frac{1}{7} )} = \frac{( 1 + \frac{1}{3} ) ( 1 + \frac{1}{5} ) ( 1 + \frac{1}{7} )}{( 1 - \frac{1}{3} ) ( 1 - \frac{1}{5} ) ( 1 - \frac{1}{7} )} = \frac{\frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7}}{\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7}} = \frac{4 \cdot 6 \cdot 8}{2 \cdot 4 \cdot 6} = \frac{8}{2} = 2 .

Final answer

B