AMC 8 Daily Practice Round 6

Complete problem set with solutions and individual problem pages

Problem 10 Easy

What is the value of \frac{2^{2}}{3^{2}-1} \times \frac{4^{2}}{5^{2}-1} \times \frac{6^{2}}{7^{2}-1} \times \cdots \times \frac{20^{2}}{21^{2}-1}?

  • A.

    \frac{1}{8}

  • B.

    \frac{1}{11}

  • C.

    \frac{1}{20}

  • D.

    \frac{1}{2}

  • E.

    \frac{1}{441}

Answer:B

By observing the expression, we notice that the denominators can be factored using the difference of squares formula (a^2 - b^2) = (a-b)(a+b).

Let's rewrite the expression accordingly:   \frac{2^{2}}{3^{2}-1} \times \frac{4^{2}}{5^{2}-1} \times \frac{6^{2}}{7^{2}-1} \times \cdots \times \frac{20^{2}}{21^{2}-1} = \frac{2^{2}}{(3-1)(3+1)} \times \frac{4^{2}}{(5-1)(5+1)} \times \frac{6^{2}}{(7-1)(7+1)} \times \cdots \times \frac{20^{2}}{(21-1)(21+1)} = \frac{2 \times 2}{2 \times 4} \times \frac{4 \times 4}{4 \times 6} \times \frac{6 \times 6}{6 \times 8} \times \cdots \times \frac{20 \times 20}{20 \times 22}

Now we can see a clear cancellation pattern emerging : = \frac{\not{2} \times 2}{\not{2} \times 4} \times \frac{\not{4} \times 4}{\not{4} \times 6} \times \frac{\not{6} \times 6}{\not{6} \times 8} \times \cdots \times \frac{\not{20} \times 20}{\not{20} \times 22}

After complete cancellation of common factors, we are left with:   = \frac{2}{4} \times \frac{4}{6} \times \frac{6}{8} \times \cdots \times \frac{20}{22}

Observing this telescoping product, we see that each numerator cancels with the denominator of the next fraction:   = \frac{2}{\not{4}} \times \frac{\not{4}}{\not{6}} \times \frac{\not{6}}{\not{8}} \times \cdots \times \frac{\not{20}}{22}

After all cancellations, only the first numerator and the last denominator remain:   = \frac{2}{22} = \frac{1}{11}

Final result: \boxed{\frac{1}{11}}

Related Problems
Problem 8 AMC 8 Daily Practice Round 6
Problem 9 AMC 8 Daily Practice Round 6
Problem 11 AMC 8 Daily Practice Round 6
Problem 12 AMC 8 Daily Practice Round 6
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