AMC 10B/12B 2025 (Problem 2/2)Jerry wrote down the ones digit of each of the first 202520252025 positive squares: 1,4,9,6,5,6,…1,4,9,6,5,6,\ldots1,4,9,6,5,6,…. What is the sum of all the numbers Jerry wrote down?(A) 9025\text{(A)}\;9025(A)9025(B) 9070\text{(B)}\;9070(B)9070(C) 9090\text{(C)}\;9090(C)9090(D) 9115\text{(D)}\;9115(D)9115(E) 9160\text{(E)}\;9160(E)9160Related TopicsCoreToolkit 75 — Try Small ExamplesToolkit 90 — Split a Sequence into Related PartsHints (4)Hint 1149656941014⋯10122232425262728292102;112122⋯192202\begin{array}{cccccccccccccccc} 1&4&9&6&5&6&9&4&1&0&&1&4&\cdots&1&0\\ 1^2&2^2&3^2&4^2&5^2&6^2&7^2&8^2&9^2&10^2&;&11^2&12^2&\cdots&19^2&20^2 \end{array}1124229326425526629724821920102;11124122⋯⋯11920202length of period=10\text{length of period}=10length of period=10Hint 21+4+9+6+5+6+9+4+1+0=451+4+9+6+5+6+9+4+1+0=451+4+9+6+5+6+9+4+1+0=45Hint 320212,…,20252⟶1,4,9,6,52021^2,\ldots,2025^2\longrightarrow1,4,9,6,520212,…,20252⟶1,4,9,6,5Hint 4Ans=45×202010+(1+4+9+6+5)\text{Ans}=45\times\frac{2020}{10}+(1+4+9+6+5)Ans=45×102020+(1+4+9+6+5)=45×202+25=9115=45\times202+25=9115=45×202+25=9115Final Answer(D) 911591159115Related Problems (7)AMC 12A 2023 (Problem 22)AMC 10A 2025 (Problem 5)AMC 10A 2025 (Problem 19)AMC 10A/12A 2025 (Problem 21/15)AMC 10B 2025 (Problem 3)AMC 10A/12A 2020 (Problem 21/19)AMC 10B 2022 (Problem 21)