Difference between revisions of "2013 AIME II Problems/Problem 3"
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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>. | + | We find that <math>T=10(1+2+\cdots +119)</math>. From Gauss's formula (or the formula for summing an arithmetic series), 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>. |
== 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}} |
Revision as of 17:26, 11 March 2014
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 (or the formula for summing an arithmetic series), we find that the value of T 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.