2022 AMC 10B Problem 5

Attempt Problem 5 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

What is the value of (1+13)(1+15)(1+17)(1−132)(1−152)(1−172)?\frac{\left(1+\frac{1}{3}\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?
3 \sqrt3
2 2
15 \sqrt{15}
4 4
105 \sqrt{105}
Answer: B
Concepts:difference of squaresradical
Small hint:
Rewrite each square-root factor as a product of two nearby fractions
Big hint:
Pair each factor in the numerator with the matching factor under the square root
Solution:
Let the given positive expression be E.E. Squaring it and using 1−1p2=(1−1p)(1+1p)1-\frac{1}{p^2}=(1-\frac{1}{p})(1+\frac{1}{p}) gives E2=∏p∈{3,5,7}1+1p1−1p=42⋅64⋅86=4. \begin{aligned} E^2&=\prod_{p\in\{3,5,7\}}\frac{1+\frac{1}{p}}{1-\frac{1}{p}}\\ &=\frac42\cdot\frac64\cdot\frac86=4. \end{aligned} Hence E=2.E=2.
Thus, the answer is B .
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