Difference between revisions of "2002 AMC 10P Problems/Problem 23"
(→Solution 1) |
m (→Solution 1) |
||
Line 31: | Line 31: | ||
\end{align*} | \end{align*} | ||
− | Notice how <math>\frac{(2^2-1^2)}{3}=1, \frac{(3^2-2^2)}{5}=1, \frac{4^2-3^2}{7}=1, \; \dots \;, \frac{1001^2-1000^2}{2001}=1</math>. This is because all of these are in the form <math>\frac{n^2-(n-1)^2}{2n-1}=\frac{n^2-(n^2-2n+1)}{2n-1}=\frac{2n-1}{2n-1}=1</math>. There are <math> | + | Notice how <math>\frac{(2^2-1^2)}{3}=1, \frac{(3^2-2^2)}{5}=1, \frac{4^2-3^2}{7}=1, \; \dots \;, \frac{1001^2-1000^2}{2001}=1</math>. This is because all of these are in the form <math>\frac{n^2-(n-1)^2}{2n-1}=\frac{n^2-(n^2-2n+1)}{2n-1}=\frac{2n-1}{2n-1}=1</math>. There are <math>1000</math> of these terms since it begins at <math>n=2</math> and ends at <math>n=1001,</math> so <math>1001-2+1=1000.</math> Therefore, <math>a-b=1000+1 - \frac{1001^2}{2003}.</math> We can either manually calculate <math>\frac{1001^2}{2003}</math> or notice that <math>\frac{1001^2}{2003} \approx \frac{1001^2}{2002}.</math> <math>\frac{1001^2}{2003} < \frac{1001^2}{2002}</math>, so <math>-\frac{1001^2}{2003} > -\frac{1001^2}{2002}.</math> Therefore, |
\begin{align*} | \begin{align*} |
Latest revision as of 12:49, 16 July 2024
Problem
Let
and
Find the integer closest to
Solution 1
Start by subtracting and and group those with a common denominator together, leaving and to the side.
Notice how . This is because all of these are in the form . There are of these terms since it begins at and ends at so Therefore, We can either manually calculate or notice that , so Therefore,
Since we can conclude that is closer to than
Thus, our answer is
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.