2022 AMC 10B Problem 5

The numbers under the square root look like the ones on top, so look for a way to connect them.

1. Split each factor under the root. A difference of squares gives $1 - \frac{1}{k^2} = \left(1 - \frac1k\right)\left(1 + \frac1k\right)$. For example,
$$1 - \frac19 = \frac89 = \frac23 \cdot \frac43 = \left(1 - \frac13\right)\left(1 + \frac13\right).$$
2. Name the two products. Let the top be $N = \left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right) = \frac43 \cdot \frac65 \cdot \frac87$, and let $M = \left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right) = \frac23 \cdot \frac45 \cdot \frac67$.
3. Rewrite the problem. By step 1, the product under the root is $N \cdot M$. So the expression is
$$\frac{N}{\sqrt{N \cdot M}} = \frac{\sqrt N \cdot \sqrt N}{\sqrt N \cdot \sqrt M} = \sqrt{\frac{N}{M}}.$$
4. Work out $N \div M$. The denominators $3, 5, 7$ are the same in both, so they cancel:
$$\frac{N}{M} = \frac{4 \cdot 6 \cdot 8}{2 \cdot 4 \cdot 6} = \frac{8}{2} = 4.$$
5. Finish. $\sqrt{4} = 2$.

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

The whole problem rests on $1 - x^2 = (1 - x)(1 + x)$. Once you see that the product under the root is "the top times its mirror image", the top cancels with half of the root and only a tiny fraction is left. Whenever you see $1 - \frac{1}{k^2}$, split it before multiplying anything out.

Plain arithmetic also works if you keep the numbers factored. The top is $\frac43 \cdot \frac65 \cdot \frac87 = \frac{192}{105} = \frac{64}{35}$. Under the root, $\frac89 \cdot \frac{24}{25} \cdot \frac{48}{49} = \frac{8 \cdot 24 \cdot 48}{9 \cdot 25 \cdot 49} = \frac{9216}{11025}$, and $\sqrt{9216} = 96$, $\sqrt{11025} = 105$, so the root is $\frac{96}{105} = \frac{32}{35}$. Then $\frac{64}{35} \div \frac{32}{35} = 2$.