AMC 10A 2020 (Problem 8)ArithmeticVery EasyWhat is the value of 1+2+3−4+5+6+7−8+⋯+197+198+199−2001+2+3-4+5+6+7-8+\cdots+197+198+199-2001+2+3−4+5+6+7−8+⋯+197+198+199−200?(A) 9800\text{(A)}\;9800(A)9800(B) 9900\text{(B)}\;9900(B)9900(C) 10000\text{(C)}\;10000(C)10000(D) 10100\text{(D)}\;10100(D)10100(E) 10200\text{(E)}\;10200(E)10200Related TopicsCoreSums of PowersCheck AnswerYour answer:ABCDECheckHints (4)Hint 11+2+3−4+5+6+7−8+⋯+197+198+199−2001+2+3-4+5+6+7-8+\cdots+197+198+199-2001+2+3−4+5+6+7−8+⋯+197+198+199−200=1+2+3+4+⋯+200−2(4+8+⋯+200)=1+2+3+4+\cdots+200-2(4+8+\cdots+200)=1+2+3+4+⋯+200−2(4+8+⋯+200)Hint 21+2+3+4+⋯+2001+2+3+4+\cdots+2001+2+3+4+⋯+200=200⋅2012=100⋅201=20100=\frac{200\cdot201}{2}=100\cdot201=20100=2200⋅201=100⋅201=20100Hint 34+8+⋯+2004+8+\cdots+2004+8+⋯+200=4(1+2+⋯+50)=4(1+2+\cdots+50)=4(1+2+⋯+50)=4⋅50⋅512=5100=4\cdot\frac{50\cdot51}{2}=5100=4⋅250⋅51=5100Hint 4Ans=1+2+3+4+⋯+200−2(4+8+⋯+200)\text{Ans}=1+2+3+4+\cdots+200-2(4+8+\cdots+200)Ans=1+2+3+4+⋯+200−2(4+8+⋯+200)=20100−2(5100)=20100−10200=9900=20100-2(5100)=20100-10200=9900=20100−2(5100)=20100−10200=9900Final Answer(B) 990099009900Related Problems (3)AMC 10B/12B 2024 (Problem 4/4)AMC 10B Fall 2021 (Problem 22)AMC 12A 2022 (Problem 16)