2002 AMC 10P Problems/Problem 23
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 | |
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All AMC 10 Problems and Solutions |
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