AMC 8 Daily Practice - Consecutive Reduction

Complete problem set with solutions and individual problem pages

Problem 1 Easy

What is the value of 100\div(100\div99)\div(99\div98) \div\dots \div(3\div2)?

  • A.

    1

  • B.

    2

  • C.

    \frac{2}{3}

  • D.

    99

  • E.

    100

Answer:B

We observe that this problem contains repeated numbers and division symbols, so we expand the parentheses and rewrite the original expression as:

100 \div (100 \div 99) \div (99 \div 98) \div (98 \div 97) \div \dots \div (3 \div 2)=100 \div 100 \times 99 \div 99 \times 98 \div 98 \times 97 \div \dots \div 3 \times 2

We notice that consecutive pairs \not{100} \div\not{100}\times\not{99} \div \not{99} \times\not{98} \div \not{98} \times \dots \div\not{3} \times 2 cancel out completely.

Therefore, the final result of the original expression is \boxed{2}.

Related Problems
Problem 2 AMC 8 Daily Practice - Consecutive Reduction
Problem 3 AMC 8 Daily Practice - Consecutive Reduction
Problem 4 AMC 8 Daily Practice - Consecutive Reduction
Problem 5 AMC 8 Daily Practice - Consecutive Reduction
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