2022 AIME II Problems/Problem 10: Difference between revisions
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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 when![\[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\]](http://latex.artofproblemsolving.com/8/d/2/8d24022c5c24f244417fc52eecd6c4daacd8d356.png)
is divided by 
.
Solution
To solve this problem, we need to use the following result:
![\[ \sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1} . \]](http://latex.artofproblemsolving.com/8/9/2/892527bd79ffaa0fb43c423064a350b1641b25b3.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
