Difference between revisions of "2022 AIME II Problems/Problem 10"
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==Solution 3== | ==Solution 3== | ||
+ | It is somewhat well know that <math>{{n \choose 2} \choose 2} = 3 {n+1 \choose 4}</math>. (In particular, this is exercise 12.3.1 in AoPS Intermediate Counting and Probability book.) | ||
+ | |||
+ | With this in mind, we can substitute out each term in the expression: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | & {{3 \choose 2} \choose 2} + {{4 \choose 2} \choose 2} + \dots + {{40 \choose 2} \choose 2} \ | ||
+ | = ~ & 3\left( {4 \choose 4} + {5 \choose 4} + \dots + {41 \choose 4} \right) \ | ||
+ | = ~ & 3 {42 \choose 5}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | We must find the remainder when this is divided by <math>1000</math>. There is no clever method: there is only bash. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | & 3 {42 \choose 5} \ | ||
+ | = ~ & \frac{42 \cdot 41 \cdot 40 \cdot 39 \cdot 38}{5 \cdot 4 \cdot 2} \ | ||
+ | = ~ & 42 \cdot 41 \cdot 39 \cdot 38 \ | ||
+ | = ~ & (40-2)(40+2)(40-1)(40+1) \ | ||
+ | = ~ & (40^2-4)(40^2-1) \ | ||
+ | = ~ & 40^4 - 5 \cdot 40^2 + 4. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | The first two terms of this are divisible by <math>1000</math>, so the remainder when the whole thing is divided by <math>1000</math> is just <math>\boxed{004}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
Since <math>40</math> seems like a completely arbitrary number, we can use Engineer's Induction by listing out the first few sums. These are, in the order of how many terms there are starting from <math>1</math> term: <math>3</math>, <math>18</math>, <math>63</math>, <math>168</math>, <math>378</math>, and <math>756</math>. Notice that these are just <math>3 \cdot \dbinom50</math>, <math>3 \cdot \dbinom61</math>, <math>3 \cdot \dbinom72</math>, <math>3 \cdot \dbinom83</math>, <math>3 \cdot \dbinom94</math>, <math>3 \cdot \dbinom{10}5</math>. It's clear that this pattern continues up to <math>38</math> terms, noticing that the "indexing" starts with <math>\dbinom32</math> instead of <math>\dbinom12</math>. Thus, the value of the sum is <math>3 \cdot \dbinom{42}{37}=2552004 \equiv \boxed{\textbf{004}} \pmod{1000}</math>. | Since <math>40</math> seems like a completely arbitrary number, we can use Engineer's Induction by listing out the first few sums. These are, in the order of how many terms there are starting from <math>1</math> term: <math>3</math>, <math>18</math>, <math>63</math>, <math>168</math>, <math>378</math>, and <math>756</math>. Notice that these are just <math>3 \cdot \dbinom50</math>, <math>3 \cdot \dbinom61</math>, <math>3 \cdot \dbinom72</math>, <math>3 \cdot \dbinom83</math>, <math>3 \cdot \dbinom94</math>, <math>3 \cdot \dbinom{10}5</math>. It's clear that this pattern continues up to <math>38</math> terms, noticing that the "indexing" starts with <math>\dbinom32</math> instead of <math>\dbinom12</math>. Thus, the value of the sum is <math>3 \cdot \dbinom{42}{37}=2552004 \equiv \boxed{\textbf{004}} \pmod{1000}</math>. | ||
Revision as of 19:01, 21 April 2022
Contents
[hide]Problem
Find the remainder whenis divided by .
Video solution
https://www.youtube.com/watch?v=4O1xiUYjnwE
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)
Solution 3
It is somewhat well know that . (In particular, this is exercise 12.3.1 in AoPS Intermediate Counting and Probability book.)
With this in mind, we can substitute out each term in the expression: We must find the remainder when this is divided by . There is no clever method: there is only bash. The first two terms of this are divisible by , so the remainder when the whole thing is divided by is just .
Solution 4
Since seems like a completely arbitrary number, we can use Engineer's Induction by listing out the first few sums. These are, in the order of how many terms there are starting from term: , , , , , and . Notice that these are just , , , , , . It's clear that this pattern continues up to terms, noticing that the "indexing" starts with instead of . Thus, the value of the sum is .
~A1001
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.