Difference between revisions of "2013 AMC 8 Problems/Problem 17"
(→Problem) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| + | The mean of these numbers is <math>\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5</math>. Therefore the numbers are <math>333, 334, 335, 336, 337, 338</math>, so the answer is <math>\boxed{\textbf{(B)}\ 338}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=16|num-a=18}} | {{AMC8 box|year=2013|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:30, 27 November 2013
Problem
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
Solution
The mean of these numbers is 
. Therefore the numbers are 
, so the answer is 
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
