Difference between revisions of "2022 AIME II Problems/Problem 10"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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+ | ==Solution 2 (similar to solution 1)== | ||
+ | |||
+ | Doing simple algebra calculation will give the following equation: | ||
+ | <cmath>\begin{align*} | ||
+ | \binom{\binom{n}{2}}{2} = \frac{\frac{n(n-1)}{2} \cdot (\frac{n(n-1)}{2}-1)}{2} | ||
+ | \\ = \frac{n(n-1)(n^2-n-2)}{8} | ||
+ | \\ = \frac{(n+1)n(n-1)(n-2)}{8} | ||
+ | \\ = \frac{(n+1)!}{8\cdot (n-3)!} = 3 \cdot \frac{(n+1)!}{4!\cdot (n-3)!} | ||
+ | \\ = 3 \binom{n+1}{4} \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Next, by using [[Hockey-Stick Identity]], we have: | ||
+ | <cmath>3 \cdot \sum_{i=3}^{40} \binom{n+1}{4} = 3 \binom{42}{5} = 42 \cdot 41 \cdot 39 \cdot 38</cmath> | ||
+ | <cmath>=(40^2-2^2)(40^2-1^2) \equiv \boxed{004} (mod 1000)</cmath> | ||
+ | |||
+ | ~DSAERF-CALMIT (https://binaryphi.site) | ||
==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 01:02, 19 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)
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)
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.