2022 AIME I Problems/Problem 6

Problem

Find the number of ordered pairs of integers $(a, b)$ such that the sequence $3, 4, 5, a, b, 30, 40, 50$ is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.

Solution 1

Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$ . Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $b$ can never be $20$ . Since $a < b$ , there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind.

However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$ . Since $3,5,a,b$ cannot form an arithmetic progression, $\underline{(a,b) \neq (7,9)}$ . $a,b,30,50$ cannot be an arithmetic progression, so $(a,b) \neq (-10,10)$ ; however, since this pair was not counted in our $231$ , we do not need to subtract it off. $3,a,b,30$ cannot form an arithmetic progression, so $\underline{(a,b) \neq (12,21)}$ . $4, a, b, 40$ cannot form an arithmetic progression, so $\underline{(a,b) \neq (16,28)}$ . $5, a,b, 50$ cannot form an arithmetic progression, $(a,b) \neq 20, 35$ ; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$ ), we do not to subtract it off.

Also, the sequences $(3,a,b,40)$ , $(3,a,b,50)$ , $(4,a,b,30)$ , $(4,a,b,50)$ , $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.

So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $\boxed{228}$ .

~ ihatemath123

Solution 2

Denote $S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}$ .

Denote by $A$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$ .

Denote by $B$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$ .

Hence, $C$ is a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$ .

Hence, this problem asks us to compute $| S | - \left( | A | + | B | + | C | \right) .$

First, we compute $| S |$ .

We have $| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276$ .

Second, we compute $| A |$ .

$\textbf{Case 1}$ : $a = 6$ .

We have $b = 8 , \cdots , 19, 21, 22, \cdots, 29$ . Thus, the number of solutions is 21.

$\textbf{Case 2}$ : $a = 20$ .

We have $b = 21, 22, \cdots , 29$ . Thus, the number of solutions is 9.

Thus, $| A | = 21 + 9 = 30$ .

Third, we compute $| B |$ .

In $B$ , we have $b = 6, 20$ . However, because $6 \leq a < b$ , we have $b \geq 7$ . Thus, $b = 20$ .

This implies $a = 7, 8, 9, 11, 12, \cdots , 19$ . Thus, $| B | = 12$ .

Fourth, we compute $| C |$ .

$\textbf{Case 1}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$ .

Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$ . Therefore, the number solutions in this case is 3.

$\textbf{Case 2}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$ .

$\textbf{Case 2.1}$ : The arithmetic sequence is $3, a, b, 30$ .

Hence, $(a, b) = (12, 21)$ .

$\textbf{Case 2.2}$ : The arithmetic sequence is $4, a, b, 40$ .

Hence, $(a, b) = (16, 28)$ .

$\textbf{Case 2.3}$ : The arithmetic sequence is $5, a, b, 50$ .

Hence, $(a, b) = (20, 35)$ .

Putting two cases together, $| C | = 6$ .

Therefore, $\begin{align*} | S | - \left( | A | + | B | + | C | \right) & = 276 - \left( 30 + 12 + 6 \right) \\ & = \boxed{\textbf{(228) }} . \end{align*}$

~Steven Chen (www.professorchenedu.com)

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2022 AIME I Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See Also