Difference between revisions of "2001 AMC 8 Problems/Problem 21"
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==Solution== | ==Solution== | ||
| − | Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is <math> 18 </math>, and there are <math> 2 </math> numbers less than <math> 18 </math> and <math> 2 </math> numbers greater than <math> 18 </math>. The sum of these integers is <math> 5(15)=75 </math>, since the mean is <math> 15 </math>. To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than <math> 18 </math> must be positive and distinct, so the smallest possible numbers for these are <math> 1 </math> and <math> 2 </math>. | + | Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is <math> 18 </math>, and there are <math> 2 </math> numbers less than <math> 18 </math> and <math> 2 </math> numbers greater than <math> 18 </math>. The sum of these integers is <math> 5(15)=75 </math>, since the mean is <math> 15 </math>. To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than <math> 18 </math> must be positive and distinct, so the smallest possible numbers for these are <math> 1 </math> and <math> 2 </math>. The number right after <math> 18 </math> also needs to be as small as possible, so it must be <math> 19 </math>. This means that the remaining number, the maximum possible value for a number in the set, is <math> 75-1-2-18-19=35, \boxed{\text{(D) 35}} </math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=20|num-a=22}} | {{AMC8 box|year=2001|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:02, 13 July 2019
Problem
The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
Solution
Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is 
, and there are 
numbers less than and numbers greater than . The sum of these integers is 
, since the mean is 
. To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than must be positive and distinct, so the smallest possible numbers for these are 
and . The number right after also needs to be as small as possible, so it must be 
. This means that the remaining number, the maximum possible value for a number in the set, is 
.
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
