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Difference between revisions of "2013 AMC 8 Problems/Problem 17"

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<math>\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350</math>
 
<math>\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350</math>
  
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==Solution==
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==Solution 1==
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The mean of these numbers is <math>\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5</math>. Therefore the numbers are <math>333, 334, 335, 336, 337, 338</math>, so the answer is <math>\boxed{\textbf{(B)}\ 338}</math>
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The arithmetic mean of these numbers is <math>\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5</math>. Therefore the numbers are <math>333</math>, <math>334</math>, <math>335</math>, <math>336</math>, <math>337</math>, <math>338</math>, so the answer is <math>\boxed{\textbf{(B)}\ 338}</math>
  
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==Alternate Algebraic Solution==
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==Solution 2==
 
Let the <math>4^{\text{th}}</math> number be <math>x</math>. Then our desired number is <math>x+2</math>.  
 
Let the <math>4^{\text{th}}</math> number be <math>x</math>. Then our desired number is <math>x+2</math>.  
  
 
Our integers are <math>x-3,x-2,x-1,x,x+1,x+2</math>, so we have that <math>6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{\textbf{(B)}\ 338}</math>.
 
Our integers are <math>x-3,x-2,x-1,x,x+1,x+2</math>, so we have that <math>6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{\textbf{(B)}\ 338}</math>.
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==Solution 3==
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Let the first term be <math>x</math>. Our integers are <math>x,x+1,x+2,x+3,x+4,x+5</math>. We have, <math>6x+15=2013\implies x=333\implies x+5=\boxed{\textbf{(B)}\ 338}</math>
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==Solution 4==
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Since there are <math>6</math> numbers, we divide <math>2013</math> by <math>6</math> to find the mean of the numbers. <math>\frac{2013}{6} = 335 \frac{1}{2}</math>.
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Then, <math>335 \frac{1}{2} + \frac{1}{2} = 336</math> (the fourth number). Fifth: <math>337</math>; Sixth: <math>\boxed {338}</math>.
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==Solution 5==
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Let the <math>6th</math> number be <math>x</math>. Then our list is: <math>x-6+x-5+x-4+x-3+x-x-1=2013</math>. Simplifying this gets you <math>6x-21=2013\implies 6x=2034</math>, which means that <math>x = \boxed{\textbf{(B)}338}</math>
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==Video Solution==
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https://youtu.be/E7DZ1oILqE0 ~savannahsolver
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== Video Solution 2 ==
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https://youtu.be/AvOXHvvTlio  Soo, DRMS, NM
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=16|num-a=18}}
 
{{AMC8 box|year=2013|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:36, 24 December 2022

Problem

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution 1

The arithmetic mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$. Therefore the numbers are $333$, $334$, $335$, $336$, $337$, $338$, so the answer is $\boxed{\textbf{(B)}\ 338}$

Solution 2

Let the $4^{\text{th}}$ number be $x$. Then our desired number is $x+2$.

Our integers are $x-3,x-2,x-1,x,x+1,x+2$, so we have that $6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{\textbf{(B)}\ 338}$.

Solution 3

Let the first term be $x$. Our integers are $x,x+1,x+2,x+3,x+4,x+5$. We have, $6x+15=2013\implies x=333\implies x+5=\boxed{\textbf{(B)}\ 338}$

Solution 4

Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\frac{2013}{6} = 335 \frac{1}{2}$. Then, $335 \frac{1}{2} + \frac{1}{2} = 336$ (the fourth number). Fifth: $337$; Sixth: $\boxed {338}$.

Solution 5

Let the $6th$ number be $x$. Then our list is: $x-6+x-5+x-4+x-3+x-x-1=2013$. Simplifying this gets you $6x-21=2013\implies 6x=2034$, which means that $x = \boxed{\textbf{(B)}338}$

Video Solution

https://youtu.be/E7DZ1oILqE0 ~savannahsolver


Video Solution 2

https://youtu.be/AvOXHvvTlio Soo, DRMS, NM


See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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