AIME II 2026 (Problem 1)Find the sum of the 10th terms of all arithmetic sequences of integers that have first term equal to 444 and include both 242424 and 343434 as terms.Related TopicsCoreArithmetic Sequences and SeriesMajorDivisibilityCheck AnswerYour answer:CheckHints (4)Hint 1Consider an arithmetic sequence4a1 →+d ⋯ →+d 24am ⋯ 34aℓ ⋯\underset{a_1}{4}\ \xrightarrow{+d}\ \cdots\ \xrightarrow{+d}\ \underset{a_m}{24}\ \cdots\ \underset{a_\ell}{34}\ \cdotsa14 +d ⋯ +d am24 ⋯ aℓ34 ⋯am=a1+(m−1)da_m=a_1+(m-1)dam=a1+(m−1)dam−a1=(m−1)d=24−4=20⇒d∣20a_m-a_1=(m-1)d=24-4=20\Rightarrow d\mid20am−a1=(m−1)d=24−4=20⇒d∣20aℓ−a1=(ℓ−1)d=34−4=30⇒d∣30a_\ell-a_1=(\ell-1)d=34-4=30\Rightarrow d\mid30aℓ−a1=(ℓ−1)d=34−4=30⇒d∣30Hint 2{d∣20d∣30 ⇒ d∣gcd(20,30)=10 ⇒ d=1,2,5,10\left\{\begin{array}{l}d\mid20\\[4pt]d\mid30\end{array}\right.\ \Rightarrow\ d\mid\gcd(20,30)=10\ \Rightarrow\ d=1,2,5,10{d∣20d∣30 ⇒ d∣gcd(20,30)=10 ⇒ d=1,2,5,10Hint 3Now compute the 10th term:a10=a1+(10−1)d=4+9da_{10}=a_1+(10-1)d=4+9da10=a1+(10−1)d=4+9dda10=a1+(10−1)d=4+9d1132225491094\begin{array}{c|c} d & a_{10}=a_1+(10-1)d=4+9d \\ \hline 1 & 13 \\ 2 & 22 \\ 5 & 49 \\ 10 & 94 \end{array}d12510a10=a1+(10−1)d=4+9d13224994Hint 4Ans=13+22+49+94=178\text{Ans}=13+22+49+94=178Ans=13+22+49+94=178Final Answer 178\text{Final Answer }178Final Answer 178Related Problems (5)AMC 10B 2022 (Problem 15)MATHCOUNTS 2026 State Target Round (Problem 5)AMC 10A 2025 (Problem 11)AMC 10B/12B 2025 (Problem 17/14)AMC 8 2026 (Problem 14)Final Answer178