2022 AMC 10B Problem 5

Let $N = \left(1+\tfrac13\right)\left(1+\tfrac15\right)\left(1+\tfrac17\right)$ be the numerator and let $M = \left(1-\tfrac13\right)\left(1-\tfrac15\right)\left(1-\tfrac17\right)$.

By the difference of squares, $1 - \tfrac{1}{k^2} = \left(1-\tfrac1k\right)\left(1+\tfrac1k\right)$, so the product under the square root is exactly $N \cdot M$. The expression becomes
$$\frac{N}{\sqrt{NM}} = \sqrt{\frac{N}{M}}.$$

Now compute the ratio factor by factor:
$$\frac{N}{M} = \frac{\tfrac43 \cdot \tfrac65 \cdot \tfrac87}{\tfrac23 \cdot \tfrac45 \cdot \tfrac67} = \frac{4 \cdot 6 \cdot 8}{2 \cdot 4 \cdot 6} = \frac{8}{2} = 4.$$

So the value is $\sqrt4 = 2$.

The answer is $\boxed{\textbf{(B)}\ 2}$.

The problem is built from the identity $1 - x^2 = (1-x)(1+x)$. Once the radicand is seen as "numerator times its mirror image," the numerator cancels against half of the root and only a tiny ratio remains. Whenever a product of $1 - \frac{1}{k^2}$ terms appears, factor it before touching a calculator.

• Multiplying out 4/3 · 6/5 · 8/7 and 8/9 · 24/25 · 48/49 numerically and then trying to simplify a huge fraction under the root.
• Forgetting the square root and answering $4$, choice (D).