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

Revision as of 01:15, 13 June 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 $1000$ .

Video solution

https://www.youtube.com/watch?v=4O1xiUYjnwE

Solution 1

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)

Solution 3

Since $40$ 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 $1$ term: $3$ , $18$ , $63$ , $168$ , $378$ , and $756$ . Notice that these are just $3 \cdot \dbinom50$ , $3 \cdot \dbinom61$ , $3 \cdot \dbinom72$ , $3 \cdot \dbinom83$ , $3 \cdot \dbinom94$ , $3 \cdot \dbinom{10}5$ . It's clear that this pattern continues up to $38$ terms, noticing that the "indexing" starts with $\dbinom32$ instead of $\dbinom12$ . Thus, the value of the sum is $3 \cdot \dbinom{42}{37}=2552004 \equiv \boxed{\textbf{004}} \pmod{1000}$ .

~A1001

@@ Line 6: / Line 6: @@
 https://www.youtube.com/watch?v=4O1xiUYjnwE
-==Solution==
+==Solution 1==
 To solve this problem, we need to use the following result:

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