Difference between revisions of "1986 AJHSME Problems/Problem 25"
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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 | 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>\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 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 \frac{50\cdot 51}{2}}{50} \\ | ||
&= \frac{2\cdot 51}{2} \\ | &= \frac{2\cdot 51}{2} \\ | ||
&= 51 \\ | &= 51 \\ | ||
| − | \end{align*}</ | + | \end{align*}</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> | 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> | ||
Revision as of 14:44, 23 May 2010
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
You can just add the first term and the last term, and see which one is the biggest. 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 highest 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 | ||
