Difference between revisions of "2013 AIME II Problems/Problem 3"
m (fixed latex) |
|||
(2 intermediate revisions by one other user not shown) | |||
Line 3: | Line 3: | ||
==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 == | == 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 00:09, 15 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.