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==Problem==
== Problem 19 ==


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?
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?
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==Solution 2==
==Solution 2==
Let <math>x</math> be the largest number. Then, <math>x+(x-2)+(x-4)+\cdots +(x-48)=10000</math>. Factoring this gives
<math>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</math>. Grouping like terms gives <math>25\left(\frac{x}{2}\right) - 300=5000</math>, and continuing down the line, we find <math>x=\boxed{\textbf{(E)}\ 424}</math>.
~MrThinker
==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>.
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>.


Note: After we combine like terms, you would have an arithmetic sequence from <math>2</math> to <math>48</math> (because <math>24 \cdot 2 = 48</math> to get last term), which would look like <math>2 + 4 + 6...+46 + 48</math> To calculate the sum of the numbers we can use the formula <math>S_n = n(\frac{a_1 + a_n}{2})</math>. This simplifies to <math>24 \cdot 25</math>, giving us <math>600</math>, which is what AfterglowBlaziken did.


~AfterglowBlaziken
~AfterglowBlaziken
~ Note by probab2023
==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 <math>\boxed{\textbf{(E)}\ 424}</math>.
~vadava_lx
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
https://youtu.be/qEq4JNouMNY
~Education, the Study of Everything


==Video Solution==
==Video Solution==
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{{AMC8 box|year=2016|num-b=18|num-a=20}}
{{AMC8 box|year=2016|num-b=18|num-a=20}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 17:16, 25 June 2025

Problem 19

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

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

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