Difference between revisions of "1993 AJHSME Problems/Problem 19"
| 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}} | ||
Revision as of 23:11, 4 July 2013
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:
![\[(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ =1800 + 1800 + 1800 + \cdots + 1800\\ =(1800)(93)\\ =\boxed{\text{(A)}\ 167,400}\]](http://latex.artofproblemsolving.com/6/a/0/6a0fbfbad183aaf6d3d02e71243a24054312abe7.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. 
