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

Revision as of 23:08, 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 $376+48 = \boxed{\textbf{(E)}\ 424}$ .

~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

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 12: / Line 12: @@
 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==

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