2022 AIME II Problems/Problem 10
Contents
Problem
Find the remainder whenis divided by .
Video Solution by OmegaLearn
https://youtu.be/pGkLAX381_s?t=1035
~ pi_is_3.14
Video solution
https://www.youtube.com/watch?v=4O1xiUYjnwE
Solution 1
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 .
~A1001
Solution 4
As in solution 1, obtain Write this as
We can safely write this expression as , since plugging and into both equal meaning they won't contribute to the sum.
Use the sum of powers formulae. We obtain
We can factor the following expression as and simplifying, we have
Substituting and simplifying gets so we would like to find To do this, get Next,
-sirswagger21
Solution 5 (Telescoping)
For the last step, see Solution 1.
~qyang
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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