AMC 8 Daily Practice - Consecutive Reduction

Complete problem set with solutions and individual problem pages

Problem 7 Medium

What is the value of \left(1+\frac{2}{3}\right)\times\left(1+\frac{2}{4}\right)\times\left(1+\frac{2}{5}\right) \times \cdots \times\left(1+\frac{2}{22}\right)?

  • A.

    46

  • B.

    24

  • C.

    22

  • D.

    \frac{12}{11}

  • E.

    \frac{5}{3}

Answer:A

Let's evaluate the expression by first converting each term to an improper fraction: \left(1+\frac{2}{3}\right) \times \left(1+\frac{2}{4}\right) \times \left(1+\frac{2}{5}\right) \times \cdots \times \left(1+\frac{2}{22}\right) = \frac{5}{3} \times \frac{6}{4} \times \frac{7}{5} \times \cdots \times \frac{24}{22}

Now we can write this product as a single fraction: = \frac{5 \times 6 \times 7 \times \cdots \times 22 \times 23 \times 24}{3 \times 4 \times 5 \times \cdots \times 22}

We observe a clear cancellation pattern: = \frac{\not{5} \times \not{6} \times \not{7} \times \cdots \times \not{22} \times 23 \times 24}{3 \times 4 \times \not{5} \times \cdots \times \not{22}}

After complete cancellation of common factors in numerator and denominator, we are left with: = \frac{23 \times 24}{3 \times 4}

Now we can simplify the remaining expression: = 23 \times \frac{24}{12} = 23 \times 2 = 46

Final result: \boxed{46}

Related Problems
Problem 4 AMC 8 Daily Practice - Consecutive Reduction
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