1986 AJHSME Problems/Problem 25

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 $\boxed{\text{D}}$

Note: 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.

1986 AJHSME (Problems • Answer Key • Resources)
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1986 AJHSME Problems/Problem 25

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