Difference between revisions of "2016 AMC 8 Problems/Problem 19"

Latest revision as of 17:34, 30 July 2024

Problem

The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?

$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$

Solution 1

Let $n$ 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 $25n$ since $(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n$ . Now, $25n=10000 \rightarrow n=400$ . Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\boxed{\textbf{(E)}\ 424}$ .

Solution 2

Let $x$ be the smallest number. The equation will become, $x+(x+2)+(x+4)+\cdots +(x+48)=10,000$ . After you combine like terms, you get $25x+(50*12)=10,000$ which turns into $10,000-600=25x$ . $25x=9400$ , so $x=376$ . Then, you add $376+48 = \boxed{\textbf{(E)}\ 424}$ .

~AfterglowBlaziken

Video Solution

https://youtu.be/NHdtjvRcDD0

~savannahsolver

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.

@@ Line 1: / Line 1: @@
-. 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?
+==Problem==
-<math>(A)\mbox{ }360\mbox{           }(B)\mbox{ }388\mbox{           }(C)\mbox{ }412\mbox{           }(D)\mbox{ }416\mbox{           }(E)\mbox{ }424\mbox{           }</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 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>.
+==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 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==
+{{AMC8 box|year=2016|num-b=18|num-a=20}}
+{{MAA Notice}}

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