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AMC 8 1993 Problem 19

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(1901+1902+1903+⋯+1993)−(101+102+103+⋯+193)=(1901+1902+1903+\cdots + 1993) - (101+102+103+\cdots + 193) = (A) 167,400(B) 172,050(C) 181,071(D) 199,300(E) 362,142\text{(A)}\ 167,400 \text{(B)}\ 172,050 \text{(C)}\ 181,071 \text{(D)}\ 199,300 \text{(E)}\ 362,142

Dataset text NuminaMath-1.5. Quoted with source.

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167,400

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