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Difference between revisions of "1986 AJHSME Problems/Problem 25"

(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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==Solution==
 
==Solution==
  
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{{Solution}}
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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.
 +
 
 +
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
 +
<cmath>\begin{align*}
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\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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&= \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\
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&= \frac{2\cdot 51}{2} \\
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&= 51 \\
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\end{align*}</cmath>
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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>
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 10:30, 24 January 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

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