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Difference between revisions of "2013 AIME II Problems/Problem 3"
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==Solution==
 
==Solution==
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We find that <math>T=10(1+2+\cdots +119)</math>. From Gauss' formula, we find that the value of T is <math>10(7140)=71400</math>. The value of <math>\frac{T}{2}</math> is therefore <math>35700</math>. We find that <math>35700</math> is <math>10(3570)=10\cdot \frac{k(k+1)}{2}</math>, so <math>3570=\frac{k(k+1)}{2}</math>. As a result, <math>7140=k(k+1)</math>, which leads to <math>0=k^2+k-7140</math>. We notice that <math>k=84</math>, so the answer is <math>10(119-84)=\boxed{350}</math>.
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We find that <math>T=10(1+2+\cdots +119)</math>. From Gauss's formula, we find that the value of <math>T</math> is <math>10(7140)=71400</math>. The value of <math>\frac{T}{2}</math> is therefore <math>35700</math>. We find that <math>35700</math> is <math>10(3570)=10\cdot \frac{k(k+1)}{2}</math>, so <math>3570=\frac{k(k+1)}{2}</math>. As a result, <math>7140=k(k+1)</math>, which leads to <math>0=k^2+k-7140</math>. We notice that <math>k=84</math>, so the answer is <math>10(119-84)=\boxed{350}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=II|num-b=2|num-a=4}}
 
{{AIME box|year=2013|n=II|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:09, 14 February 2020

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's 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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