2022 AIME II Problems/Problem 10
Contents
[hide]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 3
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 .
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 |
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