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

Revision as of 17:29, 14 January 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 largest number. Then, $x+(x-2)+(x-4)+\cdots +(x-48)=10000$ . Factoring this gives $2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 24\right)\right)=10,000$ . Grouping like terms gives $25\left(\frac{x}{2}\right) - 300=5000$ , and continuing down the line, we find $x=\boxed{\textbf{(E)}\ 424}$ .

~MrThinker

Solution 3

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

Solution 4

Dividing the series by $2$ , we get that the sum of $25$ consecutive integers is $5000$ . Let the middle number be $k$ we know that the sum is $25k$ , so $25k=5000$ . Solving, $k=200$ . $2k=400$ is the middle term of the original sequence, so the original last term is $400+\frac{25-1}{2}\cdot 2=424$ . So the answer is $\boxed{\textbf{(E)}\ 424}$ .

~vadava_lx

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/qEq4JNouMNY

~Education, the Study of Everything

Video Solution

https://youtu.be/NHdtjvRcDD0

~savannahsolver

@@ Line 23: / Line 23: @@
 ==Solution 4==
-Dividing the series by <math>2</math>, we get that the sum of <math>25</math> consecutive integers is <math>5000</math>. Let the middle number be <math>k</math> we know that the sum is <math>25k</math>, so <math>25k=5000</math>. Solving, <math>k=200</math>. <math>2k=400</math> is the middle term of the original sequence, so the original last term is <math>400+\frac{25-1}{2}\cdot 2=424</math>. So the answer is \boxed{\textbf{(E)}\ 424}$.
+Dividing the series by <math>2</math>, we get that the sum of <math>25</math> consecutive integers is <math>5000</math>. Let the middle number be <math>k</math> we know that the sum is <math>25k</math>, so <math>25k=5000</math>. Solving, <math>k=200</math>. <math>2k=400</math> is the middle term of the original sequence, so the original last term is <math>400+\frac{25-1}{2}\cdot 2=424</math>. So the answer is <math>\boxed{\textbf{(E)}\ 424}</math>.
+~vadava_lx
 ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==

2016 AMC 8 (Problems • Answer Key • Resources)
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