1993 AJHSME Problems/Problem 19

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Problem

$(1901+1902+1903+\cdots + 1993) - (101+102+103+\cdots + 193) =$

$\text{(A)}\ 167,400 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \text{(D)}\ 199,300 \qquad \text{(E)}\ 362,142$

Solution

We see that $1901=1800+101$, $1902=1800+102$, etc. Each term in the first set of numbers is $1800$ more than the corresponding term in the second set; Because there are $93$ terms in the first set, the expression can be paired up as follows and simplified:

\[(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ =1800 + 1800 + 1800 + \cdots + 1800\\ =(1800)(93)\\ =\boxed{\text{(A)}\ 167,400}\]

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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