2013 AMC 8 Problems/Problem 17: Difference between revisions

Revision as of 16:48, 25 July 2021

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 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 18: / Line 18: @@
 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>.
 Then, <math>335 \frac{1}{2} + \frac{1}{2} = 336</math> (the fourth number). Fifth: <math>337</math>; Sixth: <math>\boxed {338}</math>.
+==Solution 5==
 ==See Also==
 {{AMC8 box|year=2013|num-b=16|num-a=18}}
 {{MAA Notice}}

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