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2024 AMC 10A Problem 20

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2024 AMC 10A Problem 20AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 4 of 5 · about 7 min · position P16-20

Let $S$ be a subset of $\{1, 2, 3, \dots, 2024\}$ such that the following two conditions hold:

If $x$ and $y$ are distinct elements of $S$ , then $|x-y| > 2.$

If $x$ and $y$ are distinct odd elements of $S$ , then $|x-y| > 6.$

What is the maximum possible number of elements in $S$ ?

A)
$436$
B)
$506$
C)
$608$
D)
$654$
E)
$675$

Answer

Key insight

Ten consecutive integers hold at most 3 elements (four would be spaced exactly 3 apart, putting two odds 6 apart); the pattern 1, 4, 8 per block gives 608.

Solution

Upper bound. Take any $10$ consecutive integers. Elements of $S$ are pairwise at least $3$ apart, so at most $4$ of them lie in this stretch, and $4$ is possible only if they are $k, k+3, k+6, k+9$ . But then $k, k+6$ share a parity and so do $k+3, k+9$ ; one of these pairs consists of two odd elements only $6$ apart, contradicting the second condition. So every $10$ consecutive integers contain at most $3$ elements of $S$ .

Cut $\{1, \dots, 2024\}$ into $202$ blocks of ten ( $1$ – $10$ , …, $2011$ – $2020$ ) plus the leftover $\{2021, 2022, 2023, 2024\}$ , which holds at most $2$ elements (they must be $3$ apart). Hence $|S| \le 202\cdot 3 + 2 = 608$ .

Construction. In each block take $10k+1,\ 10k+4,\ 10k+8$ for $k = 0, 1, \dots, 201$ , and add $2021$ and $2024$ . Consecutive elements differ by $3, 4, 3, 3, 4, 3, \dots$ and finally $2018 \to 2021 \to 2024$ by $3$ , so all gaps exceed $2$ . The odd elements are exactly $1, 11, 21, \dots, 2021$ , spaced $10$ apart, which exceeds $6$ . This $S$ has $608$ elements.

The answer is $\boxed{\textbf{(C)}\ 608}$ .

Why this works

Maximum-size problems with spacing rules are solved by a density bound plus a matching periodic construction. Find the period by asking "how many elements can fit in a window of length $L$ ?" for a small $L$ ; here $L = 10$ is where the parity rule bites (four elements $3$ apart alternate parity, forcing two odds within $6$ ). Then tile the whole range with that window and handle the remainder separately.

The trap

Using the naive pattern 1, 4, 7, 10, … (odds only 6 apart) or evens spaced 4 apart (506), or dropping the extra 2 elements from the leftover 2021–2024.

Common mistakes

Using the naive pattern 1, 4, 7, 10, … (odds only 6 apart) or evens spaced 4 apart (506), or dropping the extra 2 elements from the leftover 2021–2024.
Reading $|x - y| > 2$ as "differ by at least $2$ " instead of "at least $3$ ."

Techniques

Bound the quantity above/below or estimate to pin it down · Use an invariant, parity, or coloring argument · Compute small cases, spot the pattern, generalize