Difference between revisions of "2016 AMC 8 Problems/Problem 19"
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<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math> | <math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | 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>. | + | 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>. |
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Let <math>x</math> be the smallest number. The equation will become, <math>x+(x+2)+(x+4)+\cdots +(x+48)=10,000</math>. After you combine like terms, you get <math>25x+(50*12)=10,000</math> which turns into <math>10,000-600=25x</math>. <math>25x=9400</math>, so <math>x=376</math>. Then, you add <math>376+48 = \boxed{\textbf{(E)}\ 424}</math>. | ||
| + | |||
| + | |||
| + | ~AfterglowBlaziken | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/NHdtjvRcDD0 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2016|num-b=18|num-a=20}} | {{AMC8 box|year=2016|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:34, 30 July 2024
Problem
The sum of 
consecutive even integers is 
. What is the largest of these consecutive integers?
Solution 1
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 
.
Solution 2
Let 
be the smallest number. The equation will become, 
. After you combine like terms, you get 
which turns into 
. 
, so 
. Then, you add 
.
~AfterglowBlaziken
Video Solution
~savannahsolver
See Also
| 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. 
