Difference between revisions of "1986 AJHSME Problems/Problem 25"
5849206328x (talk | contribs) (New page: ==Problem== Which of the following sets of whole numbers has the largest average? <math>\text{(A)}\ \text{multiples of 2 between 1 and 101} \qquad \text{(B)}\ \text{multiples of 3 betwee...) |
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| − | ==Problem== | + | ==Problem 25== |
Which of the following sets of whole numbers has the largest average? | Which of the following sets of whole numbers has the largest average? | ||
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==Solution== | ==Solution== | ||
| − | {{Solution}} | + | ===Solution 1=== |
| + | |||
| + | From <math>1</math> to <math>101</math> there are <math>\left\lfloor \frac{101}{2} \right\rfloor = 50</math> (see [[Floor function|floor function]]) multiples of <math>2</math>, and their average is | ||
| + | |||
| + | <center> | ||
| + | <math>\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> | ||
| + | |||
| + | Similarly, we can find that the average of the multiples of <math>3</math> between <math>1</math> and <math>101</math> is <math>51</math>, the average of the multiples of <math>4</math> is <math>52</math>, the average of the multiples of <math>5</math> is <math>52.5</math>, and the average of the multiples of <math>6</math> is <math>51</math>, so the one with the largest average is <math>\boxed{\text{D}}</math> | ||
| + | |||
| + | ===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 the 2 entries is twice the average of whole sequence). <math>2+100=102</math>, <math>3+99=102</math>, <math>4+100=104</math>, <math>5+100=105</math>, <math>6+96=102</math>. Therefore, the answer is <math>\boxed{\text{D}}</math> because it has the largest number. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/eRh_yzMOD14 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1986|num-b=24|after=Last<br>Problem}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:13, 17 January 2024
Problem 25
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 the 2 entries is twice the average of whole sequence). 
, 
, 
, 
, 
. Therefore, the answer is because it has the largest number.
Video Solution
~savannahsolver
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. 
