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

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==Solution 1==
 
==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>.
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==Solution 2==
 
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Let <math>x</math> be the largest number. Then, <math>x+(x-2)+(x-4)+\cdots +(x-48)=10000</math>. Factoring this gives
 
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<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>.
 
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~MrThinker
 
  
 
==Solution 3==
 
==Solution 3==

Revision as of 22:08, 17 May 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 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 $376+48 = \boxed{\textbf{(E)}\ 424}$.

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

~AfterglowBlaziken ~ Note by probab2023

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

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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