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

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==Problem==
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==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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<center>
 
<center>
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<math>\begin{align*}
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<math>\frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots + 2\cdot 50}{50}  
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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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\\
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&= \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\
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\\
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&= \frac{2\cdot 51}{2} \\
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= \frac{2(1+2+3+\cdots +50)}{50}  
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&= 51 \\
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\\
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\end{align*}</math>
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\\
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= \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\
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\\
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= \frac{2\cdot 51}{2} \\
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\\
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= 51 </math>
 
</center>
 
</center>
  
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===Solution 2===
 
===Solution 2===
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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.
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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.
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==Video Solution==
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https://youtu.be/eRh_yzMOD14
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{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]]
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{{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?

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

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

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

Solution

Solution 1

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

$\frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots + 2\cdot 50}{50} \\ \\ = \frac{2(1+2+3+\cdots +50)}{50} \\ \\ = \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\ \\ = \frac{2\cdot 51}{2} \\ \\ = 51$

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 $\boxed{\text{D}}$

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). $2+100=102$, $3+99=102$, $4+100=104$, $5+100=105$, $6+96=102$. Therefore, the answer is $\boxed{\text{D}}$ because it has the largest number.

Video Solution

https://youtu.be/eRh_yzMOD14

~savannahsolver

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Problem
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All AJHSME/AMC 8 Problems and Solutions

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