Difference between revisions of "1991 AJHSME Problems/Problem 19"
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The sum of nine of the numbers is <math>45</math>, and the sum of all ten is <math>100</math> so the last number must be <math>100-45=55\rightarrow \boxed{\text{C}}</math>. | The sum of nine of the numbers is <math>45</math>, and the sum of all ten is <math>100</math> so the last number must be <math>100-45=55\rightarrow \boxed{\text{C}}</math>. | ||
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==See Also== | ==See Also== | ||
Latest revision as of 11:16, 8 October 2015
Problem
The average (arithmetic mean) of 
different positive whole numbers is . The largest possible value of any of these numbers is
Solution
If the average of the numbers is , then their sum is 
.
To maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be 
. Their sum is 
.
The sum of nine of the numbers is , and the sum of all ten is 
so the last number must be 
.
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
