Difference between revisions of "2022 AIME II Problems/Problem 10"
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==Solution== | ==Solution== | ||
+ | To solve this problem, we need to use the following result: | ||
+ | |||
+ | <cmath> | ||
+ | \[ | ||
+ | \sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Now, we use this result to solve this problem. | ||
+ | |||
+ | We have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sum_{i=3}^{40} \binom{\binom{i}{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}- 2 \right)}{2} \ | ||
+ | & = \frac{1}{8} \sum_{i=3}^{40} \left( i \left( i - 1 \right) \left( i \left( i - 1 \right) - 1 \right) \right) \ | ||
+ | & = \frac{1}{8} \sum_{i=3}^{40} \left( i \left( i - 1 \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( \binom{41}{5} - \binom{3}{5} + \binom{41}{4} - \binom{3}{4} \right) \ | ||
+ | & = 3 \left( \binom{41}{5} + \binom{41}{4} \right) \ | ||
+ | & = 3 \cdot \frac{41 \cdot 40 \cdot 39 \cdot 38}{5!} \left( 37 + 5 \right) \ | ||
+ | & = 3 \cdot 41 \cdot 13 \cdot 38 \cdot 42 \ | ||
+ | & = 38 \cdot 39 \cdot 41 \cdot 42 \ | ||
+ | & = \left( 40 - 2 \right) \left( 40 - 1 \right) \left( 40 + 1 \right) \left( 40 + 2 \right) \ | ||
+ | & = \left( 40^2 - 2^2 \right) \left( 40^2 - 1^2 \right) \ | ||
+ | & = \left( 40^2 - 4 \right) \left( 40^2 - 1 \right) \ | ||
+ | & = 40^4 - 40^2 \cdot 5 + 4 . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, modulo 1000, <math>\sum_{i=3}^{40} \binom{\binom{i}{2}}{2} \equiv \boxed{\textbf{(004) }}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=9|num-a=11}} | {{AIME box|year=2022|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:58, 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.