Difference between revisions of "2002 AIME II Problems/Problem 6"
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{{AIME box|year=2002|n=II|num-b=5|num-a=7}} | {{AIME box|year=2002|n=II|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | Although the answer to Problem 6 doesn't change, the value of the telescoping sum is incorrect as given. Instead of | ||
+ | <cmath>250 \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9997}-\frac{1}{9998}-\frac{1}{9999}-\frac{1}{10000} \right), </cmath> | ||
+ | the correct sum is | ||
+ | <cmath>250 \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9999}-\frac{1}{10000}-\frac{1}{10001}-\frac{1}{10002} \right). </cmath> |
Revision as of 12:09, 15 November 2013
Problem
Find the integer that is closest to .
Solution
You know that .
So if you pull the out of the summation, you get
.
Now that telescopes, leaving you with:
is not enough to bring
lower than
so the answer is
If you didn't know , here's how you can find it out:
We know . We can use the process of fractional decomposition to split this into two fractions thus:
for some A and B.
Solving for A and B gives or
. Since there is no n term on the left hand side,
and by inspection
. Solving yields
Then we have and we can continue as before.
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.
Although the answer to Problem 6 doesn't change, the value of the telescoping sum is incorrect as given. Instead of
the correct sum is