2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 7 Easy

For how many (not necessarily positive) integer values of n is the value of 4000\cdot \left(\frac{2}{5}\right)^n an integer? (2018 AMC 10A Problem, Question#7)

  • A.

    3

  • B.

    4

  • C.

    6

  • D.

    8

  • E.

    9

Answer:E

The prime factorization of 4000 is 2^5\cdot 5^3. Therefore, the maximum number for n is 3 , and the minimum number for n is -5 . Then we must find the range from -5 to 3, which is 3-(-5)+1=8+1=\boxed{(\rm E)~9}.

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