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> | ||
| − | ==Solution== | + | ==Solution 1== |
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> | 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> | ||
| − | == | + | ==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>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | 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> | ||
==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}} | ||
Revision as of 20:35, 27 November 2013
Problem
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
Solution 1
The mean of these numbers is 
. Therefore the numbers are 
, so the answer is 
Solution 2
Let the 
number be 
. Then our desired number is 
.
Our integers are 
, so we have that 
.
Solution 3
Let the first term be . Our integers are 
. We have, 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
