Difference between revisions of "2022 AIME II Problems/Problem 10"
Mathfun1000 (talk | contribs) m (Blanked the page) (Tag: Blanking) |
|||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | Find the remainder when<cmath>\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}</cmath>is divided by <math>1000</math>. | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | ==See Also== | ||
| + | {{AIME box|year=2022|n=II|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Revision as of 07:26, 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
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. 
