2016 AMC 8 Problems/Problem 19: Difference between revisions
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<math>(A)\mbox{ }360\mbox{ }(B)\mbox{ }388\mbox{ }(C)\mbox{ }412\mbox{ }(D)\mbox{ }416\mbox{ }(E)\mbox{ }424\mbox{ }</math> | <math>(A)\mbox{ }360\mbox{ }(B)\mbox{ }388\mbox{ }(C)\mbox{ }412\mbox{ }(D)\mbox{ }416\mbox{ }(E)\mbox{ }424\mbox{ }</math> | ||
==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 simply by <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 \cdot (15-13)=424</math>. | |||
Revision as of 11:15, 23 November 2016
19. 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 simply by 
since 
. Now, 
Remembering that this is the 13th integer, we wish to find the 25th, which is 
.
