Problem:
(1901+1902+1903+⋯+1993)−(101+102+103+⋯+193)=
Answer Choices:
A. 167,400
B. 172,050
C. 181,071
D. 199,300
E. 362,142
Solution:
Each number in the first set of numbers is 1800 more than the corresponding number in the second set:
1901,1800−101,1902,1800−102,1903,1800−103,…,1993…,1800−193
Thus the sum of the first set of numbers is 93×1800=167,400 more than the sum of the second set.
Answer: A.
The problems on this page are the property of the MAA's American Mathematics Competitions