Difference between revisions of "2013 AMC 8 Problems/Problem 17"
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==Solution 5== | ==Solution 5== | ||
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> | 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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/E7DZ1oILqE0 ~savannahsolver | ||
==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 17:00, 5 May 2022
Contents
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, 
Solution 4
Since there are 
numbers, we divide 
by to find the mean of the numbers. 
.
Then, 
(the fourth number). Fifth: 
; Sixth: 
.
Solution 5
Let the 
number be . Then our list is: 
. Simplifying this gets you 
, which means that 
Video Solution
https://youtu.be/E7DZ1oILqE0 ~savannahsolver
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. 
