Difference between revisions of "2016 AMC 8 Problems/Problem 19"
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==Solution== | ==Solution== | ||
| − | Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will | + | Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to <math>25n</math> since <math>(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n</math>. Now, <math>25n=10000 \rightarrow n=400</math> Remembering that this is the 13th integer, we wish to find the 25th, which is <math>400+2(25-13)=\boxed{\textbf{(E)}\ 424}</math>. |
{{AMC8 box|year=2016|num-b=18|num-a=20}} | {{AMC8 box|year=2016|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:45, 27 November 2016
The sum of 
consecutive even integers is 
. What is the largest of these consecutive integers?
Solution
Let 
be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to 
since 
. Now, 
Remembering that this is the 13th integer, we wish to find the 25th, which is 
.
| 2016 AMC 8 (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. 
