1987 AIME Problem 12

Attempt Problem 12 of the 1987 AIME 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 1987 AIME solutions, or check the answer key.

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12.

Let mm be the smallest integer whose cube root has the form n+r,n+r, where nn is a positive integer and 0<r<11000.0<r<\frac{1}{1000}. Find n.n.

Answer: 19
Concepts:perfect powerestimationinequality
Difficulty rating: 2110
Small Hint:

For fixed n,n, the smallest possible integer is m=n3+1m=n^3+1

Big Hint:

Compare n3+1n^3+1 with (n+0.001)3(n+0.001)^3

Solution:

For a given n,n, the closest integer cube-root candidate above nn is m=n3+1.m=n^3+1. We need n3+1<(n+0.001)3,n^3+1<(n+0.001)^3, or 1<0.003n2+0.000003n+109.1<0.003n^2+0.000003n+10^{-9}. This fails at n=18n=18 and holds at n=19.n=19. The right-hand side is increasing for positive n,n, so every smaller nn fails; every larger nn has a larger least candidate m=n3+1.m=n^3+1. Hence the smallest mm occurs with n=19.n=19.

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