Difference between revisions of "2002 AMC 10P Problems/Problem 23"
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\end{align*} | \end{align*} | ||
− | Since <math>a-b>500.5,</math> we can conclude that <math>a-b</math> is closer to <math>501 | + | Since <math>a-b>500.5,</math> we can conclude that <math>a-b</math> is closer to <math>501</math> than <math>500.</math> |
Thus, our answer is <math>\boxed{\textbf{(B) } 501}.</math> | Thus, our answer is <math>\boxed{\textbf{(B) } 501}.</math> |
Revision as of 06:10, 15 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 etc. 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.