Difference between revisions of "2000 AIME I Problems/Problem 10"
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Let <math>n</math> be the value of <math>x_k</math>, <math>n+k</math> be the sum of the 99 other numbers, and <math>S</math> be the sum of <math>\textbf{all}</math> 100 numbers. Thus, <math>S = 2n + k</math>, so <math>n = \frac{S-k}{2}</math>. We then obtain the following sequence: | Let <math>n</math> be the value of <math>x_k</math>, <math>n+k</math> be the sum of the 99 other numbers, and <math>S</math> be the sum of <math>\textbf{all}</math> 100 numbers. Thus, <math>S = 2n + k</math>, so <math>n = \frac{S-k}{2}</math>. We then obtain the following sequence: |
Revision as of 22:26, 29 December 2022
Contents
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 The answer is
- JZ
- edited by erinb28lms
Solution 3(very similar to 1)
Let be the value of
,
be the sum of the 99 other numbers, and
be the sum of
100 numbers. Thus,
, so
. We then obtain the following sequence:
Solving for the left hand side, we get . This means
. Since
=
, we get
. Thus, the answer is
. ~SoilMilk
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.