MathDB
Telescoping

Source: 1993 AMC 8 Problem 19

June 28, 2011
AMC

Problem Statement

(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 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \text{(D)}\ 199,300 \qquad \text{(E)}\ 362,142