Difference between revisions of "1993 AJHSME Problems/Problem 19"
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<cmath>(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ | <cmath>(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ | ||
| − | = | + | =1800 + 1800 + \cdots + 1800\\ |
=(1800)(93)\\ | =(1800)(93)\\ | ||
=\boxed{\text{(A)}\ 167,400}</cmath> | =\boxed{\text{(A)}\ 167,400}</cmath> | ||
| Line 15: | Line 15: | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1993|num-b=18|num-a=20}} | {{AJHSME box|year=1993|num-b=18|num-a=20}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:31, 27 June 2023
Problem
Solution
We see that 
, 
, etc. Each term in the first set of numbers is 
more than the corresponding term in the second set; Because there are 
terms in the first set, the expression can be paired up as follows and simplified:
![\[(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ =1800 + 1800 + \cdots + 1800\\ =(1800)(93)\\ =\boxed{\text{(A)}\ 167,400}\]](http://latex.artofproblemsolving.com/9/8/5/9859b07f48b50dca996860593e380896aadc0c3f.png)
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. 
