AMC 8 Daily Practice Round 6

Complete problem set with solutions and individual problem pages

Problem 7 Easy

What is the value of \frac{1}{2} \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times \cdots \times\left(1-\frac{1}{100}\right)?

  • A.

    \frac{1}{100}

  • B.

    \frac{1}{2}

  • C.

    \frac{1}{3}

  • D.

    \frac{1}{4}

  • E.

    \frac{1}{99}

Answer:A

First, we calculate the value inside each parenthesis, so the original expression can be rewritten as: \frac{1}{2} \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) \times \cdots \times \left(1 - \frac{1}{100}\right)=\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{99}{100}

We observe a telescoping pattern where the denominator of each fraction cancels with the numerator of the subsequent fraction: \frac{1}{\not{2}} \times \frac{\not{2}}{\not{3}} \times \frac{\not{3}}{\not{4}} \times \cdots \times \frac{\not{99}}{100}

After complete cancellation of adjacent terms, only the first numerator and the last denominator remain: = \frac{1}{100}

Final result: \boxed{\frac{1}{100}}

Related Problems
Problem 5 AMC 8 Daily Practice Round 6
Problem 6 AMC 8 Daily Practice Round 6
Problem 8 AMC 8 Daily Practice Round 6
Problem 9 AMC 8 Daily Practice Round 6
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