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

Latest revision as of 23:28, 23 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 $376+48 = \boxed{\textbf{(E)}\ 424}$ .

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

@@ Line 1: / Line 1: @@
+==Problem==
 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?
 <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>
-==Solution==
+==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 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+25-13=\textbf{(C)}412</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>.
+==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>.
+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
+~ 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==
+https://youtu.be/NHdtjvRcDD0
+~savannahsolver
+==See Also==
 {{AMC8 box|year=2016|num-b=18|num-a=20}}
 {{MAA Notice}}

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