Difference between revisions of "2000 AIME I Problems/Problem 10"
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==Video solution== | ==Video solution== |
Revision as of 18:20, 20 November 2022
Problem
A sequence of numbers has the property that, for every integer
between
and
inclusive, the number
is
less than the sum of the other
numbers. Given that
where
and
are relatively prime positive integers, find
.
Solution
Let the sum of all of the terms in the sequence be . Then for each integer
,
. Summing this up for all
from
,
Now, substituting for , we get
, and the answer is
.
Solution 2
Consider and
. Let
be the sum of the rest 98 terms. Then
and
Eliminating
we have
So the sequence is arithmetic with common difference
In terms of the sequence is
Therefore
. Solving, we get
75+98=\boxed{173.}$
-JZ
- minor edit by erinb28lms (final solution was not given)
Video solution
https://www.youtube.com/watch?v=TdvxgrSZTQw
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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