Difference between revisions of "2022 AIME II Problems/Problem 10"
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& = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \ | & = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \ | ||
& = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \ | & = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \ | ||
− | & = \frac{1}{8} \sum_{i=3}^{40} | + | & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) \left( i \left( i - 1 \right) - 2 \right) \ |
− | & = \frac{1}{8} \sum_{i=3}^{40} | + | & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) |
\left( \left( i - 2 \right) \left( i - 3 \right) + 4 \left( i - 2 \right) | \left( \left( i - 2 \right) \left( i - 3 \right) + 4 \left( i - 2 \right) | ||
− | \right) | + | \right) \ |
& = 3 \left( \sum_{i=3}^{40} \binom{i}{4} + \sum_{i=3}^{40} \binom{i}{3} \right) \ | & = 3 \left( \sum_{i=3}^{40} \binom{i}{4} + \sum_{i=3}^{40} \binom{i}{3} \right) \ | ||
& = 3 \left( \binom{41}{5} - \binom{3}{5} + \binom{41}{4} - \binom{3}{4} \right) \ | & = 3 \left( \binom{41}{5} - \binom{3}{5} + \binom{41}{4} - \binom{3}{4} \right) \ |
Revision as of 12:01, 18 February 2022
Problem
Find the remainder whenis divided by .
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)
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.