2026 AIME II Problem 1

Attempt Problem 1 of the 2026 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2026 AIME II solutions, or check the answer key.

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1.

Find the sum of the 1010th terms of all arithmetic sequences of integers that have first term equal to 44 and include both 2424 and 3434 as terms.

Answer: 178
Concepts:arithmetic sequencedivisibilitygreatest common divisor
Difficulty rating: 1840
Solution:

Let the common difference be d.d. Since the first term is 44 and both 2424 and 3434 appear, dd divides 244=2024 - 4 = 20 and 344=30,34 - 4 = 30, so dd divides gcd(20,30)=10.\gcd(20, 30) = 10. The difference must be positive to reach 2424 and 3434 from 4,4, so d{1,2,5,10}d \in \{1, 2, 5, 10\} (and each of these works, since d20d \mid 20 and d30d \mid 30 put both targets in the sequence).

The 1010th term is 4+9d,4 + 9d, so the requested sum is d{1,2,5,10}(4+9d)=44+9(1+2+5+10)=16+162=178. \begin{aligned} &\sum_{d \in \{1,2,5,10\}} (4 + 9d) \\ &= 4 \cdot 4 + 9(1 + 2 + 5 + 10) \\ &= 16 + 162 = 178. \end{aligned}

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