2007 AMC 10A Problem 17

Attempt Problem 17 of the 2007 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

Suppose that mm and nn are positive integers such that 75m=n3.75m = n^3. What is the minimum possible value of m+n?m + n?
1515
3030
5050
6060
57005700
Answer: D
Concepts:perfect powerprime factorization
Small hint:
In a perfect cube, every prime’s exponent is a multiple of 33
Big hint:
Since 75=3⋅52,75 = 3 \cdot 5^2, find the smallest mm making all exponents multiples of 33
Solution:
Since n3=75m=3⋅52⋅m,n^3 = 75m = 3 \cdot 5^2 \cdot m, every prime factor must occur a multiple of three times.
The smallest such mm is 32⋅5=45,3^2 \cdot 5 = 45, giving n3=33⋅53n^3 = 3^3 \cdot 5^3 and n=15.n = 15.
Then m+n=45+15=60.m + n = 45 + 15 = 60.
Thus, the correct answer is D.
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