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Difference between revisions of "2022 AIME II Problems/Problem 10"

(→‎Solution)
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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*}
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\binom{\binom{n}{2}}{2} = \frac{\frac{n(n-1)}{2} \cdot (\frac{n(n-1)}{2}-1)}{2}
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\\ = \frac{n(n-1)(n^2-n-2)}{8}
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\\ = \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)!}
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\\ = 3 \binom{n+1}{4} \end{align*}
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</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 00:02, 19 February 2022

Problem

Find the remainder when\[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\]is divided by $1000$.

Solution

To solve this problem, we need to use the following result:

\[ \sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1} . \]

Now, we use this result to solve this problem.

We have \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}- 1 \right)}{2} \\ & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) \left( i \left( i - 1 \right) - 2 \right) \\ & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) \left( \left( i - 2 \right) \left( i - 3 \right) + 4 \left( i - 2 \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*}

Therefore, modulo 1000, $\sum_{i=3}^{40} \binom{\binom{i}{2}}{2} \equiv \boxed{\textbf{(004) }}$.

~Steven Chen (www.professorchenedu.com)

Solution 2 (similar to solution 1)

Doing simple algebra calculation will give the following equation: \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*}

Next, by using Hockey-Stick Identity, we have: \[3 \cdot \sum_{i=3}^{40} \binom{n+1}{4} = 3 \binom{42}{5} = 42 \cdot 41 \cdot 39 \cdot 38\] \[=(40^2-2^2)(40^2-1^2) \equiv \boxed{004} (mod 1000)\]

~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

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