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Difference between revisions of "1993 AJHSME Problems/Problem 19"

(Created page with "==Problem== <math>(1901+1902+1903+\cdots + 1993) - (101+102+103+\cdots + 193) = </math> <math>\text{(A)}\ 167,400 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \...")
 
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<math>\text{(A)}\ 167,400 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \text{(D)}\ 199,300 \qquad \text{(E)}\ 362,142</math>
 
<math>\text{(A)}\ 167,400 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \text{(D)}\ 199,300 \qquad \text{(E)}\ 362,142</math>
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==Solution==
 +
We see that <math>1901=1800+101</math>, <math>1902=1800+102</math>, etc. Each term in the first set of numbers is <math>1800</math> more than the corresponding term in the second set; Because there are <math>93</math> terms in the first set, the expression can be paired up as follows and simplified:
 +
 +
<cmath>(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\
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=1800 + 1800 + \cdots + 1800\\
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=(1800)(93)\\
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=\boxed{\text{(A)}\ 167,400}</cmath>
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==See Also==
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{{AJHSME box|year=1993|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 11:31, 27 June 2023

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:

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

See Also

1993 AJHSME (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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