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Difference between revisions of "1986 AJHSME Problems/Problem 25"
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There seems to be no better way to solve this other than just find each of those, so that's what we do.
 
There seems to be no better way to solve this other than just find each of those, so that's what we do.
  
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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]]) multiples of <math>2</math>, and their average is  
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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  
 
<cmath>\begin{align*}
 
<cmath>\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} \\
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==See Also==
 
==See Also==
  
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[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=24|after=Last<br>Problem}}
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[[Category:Introductory Algebra Problems]]

Revision as of 18:17, 23 May 2009

Problem

Which of the following sets of whole numbers has the largest average?

$\text{(A)}\ \text{multiples of 2 between 1 and 101} \qquad \text{(B)}\ \text{multiples of 3 between 1 and 101}$

$\text{(C)}\ \text{multiples of 4 between 1 and 101} \qquad \text{(D)}\ \text{multiples of 5 between 1 and 101}$

$\text{(E)}\ \text{multiples of 6 between 1 and 101}$

Solution

There seems to be no better way to solve this other than just find each of those, so that's what we do.

From $1$ to $101$ there are $\left\lfloor \frac{101}{2} \right\rfloor = 50$ (see floor function) multiples of $2$, 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*}

Similarly, we can find that the average of the multiples of $3$ between $1$ and $101$ is $51$, the average of the multiples of $4$ is $52$, the average of the multiples of $5$ is $52.5$, and the average of the multiples of $6$ is $51$, so the one with the largest average is $\boxed{\text{D}}$

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
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