Difference between revisions of "2022 AIME II Problems/Problem 10"
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==Solution 4== | ==Solution 4== |
Revision as of 20:02, 21 April 2022
Contents
Problem
Find the remainder whenis divided by
.
Video solution
https://www.youtube.com/watch?v=4O1xiUYjnwE
Solution
To solve this problem, we need to use the following result:
Now, we use this result to solve this problem.
We have
Therefore, modulo 1000, .
~Steven Chen (www.professorchenedu.com)
Solution 2 (similar to solution 1)
Doing simple algebra calculation will give the following equation:
Next, by using Hockey-Stick Identity, we have:
~DSAERF-CALMIT (https://binaryphi.site)
Solution 4
Since seems like a completely arbitrary number, we can use Engineer's Induction by listing out the first few sums. These are, in the order of how many terms there are starting from
term:
,
,
,
,
, and
. Notice that these are just
,
,
,
,
,
. It's clear that this pattern continues up to
terms, noticing that the "indexing" starts with
instead of
. Thus, the value of the sum is
.
~A1001
See Also
2022 AIME II (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.