Difference between revisions of "2016 AMC 8 Problems/Problem 19"
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− | + | ==Problem== | |
− | <math> | + | The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers? |
− | ==Solution== | + | <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> |
− | 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 | + | |
+ | ==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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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== | ||
{{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 23:10, 17 May 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 3
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.