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AIME 2013 II · 第 3 题

AIME 2013 II — Problem 3

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem 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\tfrac{T}{2} seconds after it is lit, the candle's height in centimeters will be hh. Find 10h10h.

解析

Solution

We find that T=10(1+2++119)T=10(1+2+\cdots +119). From Gauss's formula, we find that the value of TT is 10(7140)=7140010(7140)=71400. The value of T2\frac{T}{2} is therefore 3570035700. We find that 3570035700 is 10(3570)=10k(k+1)210(3570)=10\cdot \frac{k(k+1)}{2}, so 3570=k(k+1)23570=\frac{k(k+1)}{2}. As a result, 7140=k(k+1)7140=k(k+1), which leads to 0=k2+k71400=k^2+k-7140. We notice that k=84k=84, so the answer is 10(11984)=35010(119-84)=\boxed{350}.