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2018 AMC 12A Problem 7

Problem 7 of 25EasierNumber TheoryArithmetic

For how many (not necessarily positive) integer values of nn is the value of 4000⋅(25)n 4000 \cdot \left(\tfrac{2}{5}\right)^n an integer?

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Solution

Since 4000=25⋅53,4000 = 2^5 \cdot 5^3, the expression equals 25+n⋅53−n.2^{5+n} \cdot 5^{3-n}. This is an integer exactly when both 5+n≥05 + n \ge 0 and 3−n≥0,3 - n \ge 0, that is, −5≤n≤3.-5 \le n \le 3. There are 3−(−5)+1=93 - (-5) + 1 = 9 such integers. Thus, the correct answer is E.
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Tagged: prime factorization · exponent

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