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Difference between revisions of "2013 AIME II Problems/Problem 3"
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==Problem 3==
 
A large candle is <math>119</math> 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 <math>10</math> seconds to burn down the first centimeter from the top, <math>20</math> seconds to burn down the second centimeter, and <math>10k</math> seconds to burn down the <math>k</math>-th centimeter.  Suppose it takes <math>T</math> seconds for the candle to burn down completely.  Then <math>\tfrac{T}{2}</math> seconds after it is lit, the candle's height in centimeters will be <math>h</math>.  Find <math>10h</math>.
 
A large candle is <math>119</math> 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 <math>10</math> seconds to burn down the first centimeter from the top, <math>20</math> seconds to burn down the second centimeter, and <math>10k</math> seconds to burn down the <math>k</math>-th centimeter.  Suppose it takes <math>T</math> seconds for the candle to burn down completely.  Then <math>\tfrac{T}{2}</math> seconds after it is lit, the candle's height in centimeters will be <math>h</math>.  Find <math>10h</math>.
  

Revision as of 15:50, 6 April 2013

Problem 3

A large candle is $119$ 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 $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$-th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$. Find $10h$.

Solution

We find that $T=10(1+2+\cdots +119)$. From Gauss' formula, we find that the value of T is $10(7140)=71400$. The value of $\frac{T}{2}$ is therefore $35700$. We find that $35700$ is $10(3570)=10\cdot \frac{k(k+1)}{2}$, so $3570=\frac{k(k+1)}{2}$. As a result, $7140=k(k+1)$, which leads to $0=k^2+k-7140$. We notice that $k=84$, so the answer is $10(119-84)=\boxed{350}$.

See also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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