AMC 10B 2025 (Problem 18)What is the ones digit of the sum⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊2025⌋?\lfloor\sqrt1\rfloor+\lfloor\sqrt2\rfloor+\lfloor\sqrt3\rfloor+\cdots+\lfloor\sqrt{2025}\rfloor?⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊2025⌋?(Recall that ⌊x⌋\lfloor x\rfloor⌊x⌋ represents the greatest integer less than or equal to xxx.)(A) 1\text{(A)}\;1(A)1(B) 2\text{(B)}\;2(B)2(C) 3\text{(C)}\;3(C)3(D) 5\text{(D)}\;5(D)5(E) 8\text{(E)}\;8(E)8Related TopicsCoreToolkit 75 — Try Small ExamplesToolkit 2 — Sum of squaresMinorToolkit 47 — Modular Arithmetic: Definition and PropertiesHints (3)Hint 1⌊1⌋=⌊2⌋=⌊3⌋=1\lfloor\sqrt1\rfloor=\lfloor\sqrt2\rfloor=\lfloor\sqrt3\rfloor=1⌊1⌋=⌊2⌋=⌊3⌋=1⌊4⌋=⌊5⌋=⌊6⌋=⌊7⌋=⌊8⌋=2\lfloor\sqrt4\rfloor=\lfloor\sqrt5\rfloor=\lfloor\sqrt6\rfloor=\lfloor\sqrt7\rfloor=\lfloor\sqrt8\rfloor=2⌊4⌋=⌊5⌋=⌊6⌋=⌊7⌋=⌊8⌋=2⌊9⌋=⋯=⌊15⌋=3\lfloor\sqrt9\rfloor=\cdots=\lfloor\sqrt{15}\rfloor=3⌊9⌋=⋯=⌊15⌋=3Hint 2⌊k2⌋=⌊k2+1⌋=⋯=⌊(k+1)2−1⌋=k\lfloor\sqrt{k^2}\rfloor=\lfloor\sqrt{k^2+1}\rfloor=\cdots=\left\lfloor\sqrt{(k+1)^2-1}\right\rfloor=k⌊k2⌋=⌊k2+1⌋=⋯=⌊(k+1)2−1⌋=kHint 32025=4522025=45^22025=452⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊2025⌋\lfloor\sqrt1\rfloor+\lfloor\sqrt2\rfloor+\lfloor\sqrt3\rfloor+\cdots+\lfloor\sqrt{2025}\rfloor⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊2025⌋=(⌊1⌋+⌊2⌋+⌊3⌋)+(⌊4⌋+⋯+⌊8⌋)=(\lfloor\sqrt1\rfloor+\lfloor\sqrt2\rfloor+\lfloor\sqrt3\rfloor)+(\lfloor\sqrt4\rfloor+\cdots+\lfloor\sqrt8\rfloor)=(⌊1⌋+⌊2⌋+⌊3⌋)+(⌊4⌋+⋯+⌊8⌋)+(⌊9⌋+⋯+⌊15⌋)+⋯+(⌊k2⌋+⋯+⌊(k+1)2−1⌋)+(\lfloor\sqrt9\rfloor+\cdots+\lfloor\sqrt{15}\rfloor)+\cdots+(\lfloor\sqrt{k^2}\rfloor+\cdots+\lfloor\sqrt{(k+1)^2-1}\rfloor)+(⌊9⌋+⋯+⌊15⌋)+⋯+(⌊k2⌋+⋯+⌊(k+1)2−1⌋)+⋯+(⌊442⌋+⋯+⌊452−1⌋)+⌊2025⌋+\cdots+(\lfloor\sqrt{44^2}\rfloor+\cdots+\lfloor\sqrt{45^2-1}\rfloor)+\lfloor\sqrt{2025}\rfloor+⋯+(⌊442⌋+⋯+⌊452−1⌋)+⌊2025⌋=3×1+5×2+7×3+⋯+(2k+1)k+⋯+89×44+45=3\times1+5\times2+7\times3+\cdots+(2k+1)k+\cdots+89\times44+45=3×1+5×2+7×3+⋯+(2k+1)k+⋯+89×44+45=∑k=144(2k2+k)+45=\sum_{k=1}^{44}(2k^2+k)+45=k=1∑44(2k2+k)+45=2(12+22+⋯+442)+(1+2+⋯+44)+45=2(1^2+2^2+\cdots+44^2)+(1+2+\cdots+44)+45=2(12+22+⋯+442)+(1+2+⋯+44)+45=2⋅44⋅45⋅896+44⋅452+45=2\cdot\frac{44\cdot45\cdot89}{6}+\frac{44\cdot45}{2}+45=2⋅644⋅45⋅89+244⋅45+45=2⋅22⋅15⋅89+45⋅23=2\cdot22\cdot15\cdot89+45\cdot23=2⋅22⋅15⋅89+45⋅23≡2⋅2⋅5⋅9+5⋅3≡0+5≡5(mod10)\equiv2\cdot2\cdot5\cdot9+5\cdot3\equiv0+5\equiv5\pmod{10}≡2⋅2⋅5⋅9+5⋅3≡0+5≡5(mod10)Final Answer(D) 555Related Problems (8)AMC 12A 2023 (Problem 22)AMC 10A 2025 (Problem 19)AMC 10A/12A 2025 (Problem 21/15)AMC 10B/12B 2025 (Problem 2/2)AMC 10B 2025 (Problem 3)AMC 10B/12B 2025 (Problem 17/14)AMC 10A/12A 2020 (Problem 21/19)AMC 10B 2022 (Problem 21)