AMC 8 Daily Practice Round 6

Complete problem set with solutions and individual problem pages

Problem 9 Easy

What id the value of \left(1-\frac{1}{2}\right) \times\left(2-\frac{2}{3}\right) \times\left(3-\frac{3}{4}\right) \times \cdots \times\left(8-\frac{8}{9}\right) \times\left(9-\frac{9}{10}\right)?

  • A.

    \frac{1}{2}

  • B.

    \frac{1}{3}

  • C.

    \frac{1}{10}

  • D.

    362880

  • E.

    36288

Answer:E

 Observing the expression, we notice repeated numbers in each parenthesis.

Factoring out these common terms, we can rewrite the expression as:   1 \times\left(1-\frac{1}{2}\right) \times 2 \times \left(1-\frac{1}{3}\right) \times 3 \times \left(1-\frac{1}{4}\right) \times \cdots \times 8 \times \left(1-\frac{1}{9}\right) \times 9 \times \left(1-\frac{1}{10}\right)

Rearranging the terms by moving the constants to the left and the fractions to the right, we get:   [1 \times 2 \times 3 \times \cdots \times 9] \times \left[\left(1-\frac{1}{2}\right) \times \left(1-\frac{1}{3}\right) \times \cdots \times \left(1-\frac{1}{10}\right)\right]

The right part of the expression is a telescoping product:   \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \cdots \times \left(\frac{9}{10}\right) = \frac{1}{\not{2}} \times \frac{\not{2}}{\not{3}} \times \frac{\not{3}}{\not{4}} \times \cdots \times \frac{\not{9}}{10} = \frac{1}{10}

Therefore, the original expression simplifies to:   9! \times \frac{1}{10} = \frac{9!}{10} = \frac{362880}{10} = 36288

Final result: \boxed{36288}

Related Problems
Problem 7 AMC 8 Daily Practice Round 6
Problem 8 AMC 8 Daily Practice Round 6
Problem 10 AMC 8 Daily Practice Round 6
Problem 11 AMC 8 Daily Practice Round 6
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