2013 AIME II Problem 3

Below is the professionally curated solution for Problem 3 of the 2013 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME II solutions, or check the answer key.

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Concepts:arithmetic sequencesummationfactoring

Difficulty rating: 1970

3.

A large candle is 119119 centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes 1010 seconds to burn down the first centimeter from the top, 2020 seconds to burn down the second centimeter, and 10k10k seconds to burn down the kk-th centimeter. Suppose it takes TT seconds for the candle to burn down completely. Then T2\frac{T}{2} seconds after it is lit, the candle's height in centimeters will be h.h. Find 10h.10h.

Solution:

Burning the first xx centimeters takes 10(1+2++x)=5x(x+1)10(1 + 2 + \cdots + x) = 5x(x+1) seconds, so T=5119120=71400T = 5 \cdot 119 \cdot 120 = 71400 and T2=35700.\frac{T}{2} = 35700.

Setting 5x(x+1)=357005x(x+1) = 35700 gives x(x+1)=7140=8485,x(x+1) = 7140 = 84 \cdot 85, so at time T2\frac{T}{2} the candle has burned down exactly 8484 centimeters. Its height is h=11984=35,h = 119 - 84 = 35, and 10h=350.10h = 350.

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