Difference between revisions of "1986 AJHSME Problems/Problem 25"
m (Edited Solution 2 for Clarity) |
Aquakitty11 (talk | contribs) |
||
| Line 33: | Line 33: | ||
{{AJHSME box|year=1986|num-b=24|after=Last<br>Problem}} | {{AJHSME box|year=1986|num-b=24|after=Last<br>Problem}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:35, 3 July 2013
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
$\begin{align*} \frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots + 2\cdot 50}{50} &= \frac{2(1+2+3+\cdots +50)}{50} \\ &= \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\ &= \frac{2\cdot 51}{2} \\ &= 51 \\ \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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. 
