Difference between revisions of "1986 AJHSME Problems/Problem 25"
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− | <math> | + | <math>\frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots + 2\cdot 50}{50} |
− | \frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots + 2\cdot 50}{50} | + | \\ |
− | + | \\ | |
− | + | = \frac{2(1+2+3+\cdots +50)}{50} | |
− | + | \\ | |
− | + | \\ | |
+ | = \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\ | ||
+ | \\ | ||
+ | = \frac{2\cdot 51}{2} \\ | ||
+ | \\ | ||
+ | = 51 </math> | ||
</center> | </center> | ||
Revision as of 12:10, 5 June 2015
Problem
Which of the following sets of whole numbers has the largest average?
Solution
Solution 1
From to there are (see floor function) multiples of , and their average is
Similarly, we can find that the average of the multiples of between and is , the average of the multiples of is , the average of the multiples of is , and the average of the multiples of is , so the one with the largest average is
Solution 2
The multiples of any number in any range form an arithmetic sequence. It can be proven that the average of the numbers in an arithmetic sequence is simply the average of their highest and lowest entries, so you can just add the first term and the last term, and see which one is the largest (since the sum of two numbers is twice their average). 2+100=102, 3+99=102, 4+100=104, 5+100=105, 6+96=102. Therefore, the answer is multiples of five because it has the largest number.
See Also
1986 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.