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>. | ||
==Solution== | ==Solution== | ||
| − | 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>. | + | 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 == | ||
| + | {{AIME box|year=2013|n=II|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:09, 14 February 2020
Problem 3
A large candle is 
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 
seconds to burn down the first centimeter from the top, 
seconds to burn down the second centimeter, and 
seconds to burn down the 
-th centimeter. Suppose it takes 
seconds for the candle to burn down completely. Then 
seconds after it is lit, the candle's height in centimeters will be 
. Find 
.
Solution
We find that 
. From Gauss's formula, we find that the value of is 
. The value of 
is therefore 
. We find that is 
, so 
. As a result, 
, which leads to 
. We notice that 
, so the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
