# About Average

Can you find sets of positive integers that satisfy the following?

1. Three numbers with mean $3$ and mode $2$

2. Three numbers with mean $7$ and mode $10$

3. Three numbers with mean $8$, median $10$ and range $8$

4. Four numbers with mean $7.5$, mode $6$ and median $7$

5. Four numbers with mean $6$, median $6.5$ and range $11$

6. Five numbers with mean $4$, mode $3$ and range $9$

7. Five numbers with mean $4$, mode $2$ and range $6$ (two possible solutions)

8. Five numbers with mean $7$, mode $7$ and range $10$ (three possible solutions)**Extension**

Can you find a set of four numbers with mean $4$, mode $1$, median $2$ and range $10$?

How about a set of five numbers?

Or six numbers?

Or...

Or 100 numbers?

*If you enjoyed this problem, you may also be interested in M, M and M.**With thanks to Don Steward, whose ideas formed the basis of this problem.*

Giulio from St Joseph's Catholic Primary School, Etienne, Youcef, Anna and Lily from Strand on the Green Junior School and Rakhvin from St Stephen's School Carramar in Australia all used the following method for the first two sets. Here is Rakhvin's solution:

1. Knowing that the mode was $2$, two of the three numbers must be $2$. Next, as the mean was $3$ and there were three numbers, I multiplied to get a total of $9$ for all three numbers. From there, I could do $9-2-2$ to find the last number, $5$.

2. Knowing that the mode was $10$, two of the three numbers must be $10$. Next, as the mean was $7$ and there were three numbers, I multiplied to get a total of $21$ for all three numbers. From there, I could do $21-10-10$ to find the last number, $1$.

For the third and fourth questions, these students again used similar methods. Here are the solutions of Etienne, Youcef, Anna and Lily:

3. The median of $10$ tells us that this is the middle number.

The mean of $8$ tells us that all three numbers must add up to $24$, so the other two numbers must add up to $14$. Since the range is $8$, they have a difference of $8$, so the numbers must be $3$ and $11$.

Therefore the overall set is $3$, $10$ and $11$.

4. The mode of $6$ means that there are at least two $6$'s.

For the median to be $7$, the other middle number is $8$. For the mean to be $7.5$, the total must be $4 \times 7.5 = 30$, and therefore the final number is $10$.

Jacob from Sacred Heart Catholic College, as well as Etienne, Youcef, Anna and Lily, gave the following explanation as to why the description given in question 5 has no solutions. Here is Jacob's solution:

It is impossible to find a set of four positive integers that suit these criteria. First of all, the set of numbers must include two middle numbers that add up to $13$ in order to get a median of $6.5$.

We know that all four numbers in the set must add up to $24$ because the mean is $6$ and $6$ multiplied by $4$ equals $24$. So far the two numbers that we have add up to $13$. In addition to this, the four numbers must have a range of $11$. This means that the other two numbers, which are the first and last numbers in the sequence, must add up to $11$. The only two numbers that can be used are $0$ and $11$.

However, $0$ is not a positive integer; it is neither a positive or negative integer. Due to the fact that we cannot use $0$ as one of the numbers in the set of four positive integers, it is impossible to get a set of four numbers at all. This is because there are no other sets of numbers that fulfil this criteria.

However, other students noticed that if you were to allow $0$, then there would be solutions. For example, Rakhvin obtained the set:

Liam, Etienne, Youcef, Anna and Lily also gave solutions for questions 6, 7 and 8:

7. A mode of $2$ means that there are at least two $2$'s. A mean of $4$ means that the other three numbers add up to $16$.

If the lowest integer is $1$, the range of $6$ tells us that the largest is $7$ but we would need an $8$ to make them add up to $16$, so that doesn't work. So the smallest number is $2$, the highest is $8$ (from the range) and the other two number can be $2$ and $6$ or $3$ and $5$. They can't be $4$ and $4$ because then the numbers would have two modes.

This gives overall the sets: $2,2,2,6,8$ and $2,2,3,5,8$.

8. The mode of $7$ means there are at least two $7$'s. The mean of $7$ tells us that the other three numbers add up to $21$. The range is $10$, so if $1$ is the smallest number, the largest is $11$ and that leaves $9$. If $2$ is the smallest number, the largest is $12$ and that leaves $7$. If $3$ is the smallest number, the largest is $13$ and that leaves $5$. If $4$ is the smallest number, the largest is $14$ and that leaves only $3$, but that can't be the case because 4 had to be the smallest number.

So the smallest number has to be 3 or less.

Therefore, the answers are: $1,7,7,9,11$, $2,7,7,7,12$ and $3,5,7,7,13$.

These solutions can all also be done using algebra to help. Both Mitsuyo from Garden International School in Kuala Lumpur, Malaysia and Zach used this method. You can find Mitsuyo's solution here and Zach's solution here .

Four numbers: $1,1,3,11$

Five numbers: $1,1,2,5,11$

Six numbers: $1,1,1,3,7,11$

Class 8a2 from Ousedale School were able to find a pattern that would always work:

4 numbers: $1,1,3,11$

6 numbers: $1,1,1,3,7,11$

8 numbers: $1,1,1,1,3,7,7,11$

10 numbers:$1,1,1,1,1,3,7,7,7,11$

12 numbers:$1,1,1,1,1,1,3,7,7,7,7,11$

They could then generalise this using algebra:

With $n$ numbers, where $n$ is even, there are:

$\frac{n}{2}$ ones, $1$ three, $\frac{n}{2} -2$ sevens and $1$ eleven.

Great stuff! Can anyone find a generalisation for when $n$ is odd?

**Why do this problem?**

This problem improves fluency in working with the numerical definitions of the mean, median, mode and range. It also develops understanding of how each measure is affected by individual numbers in a sample, as well as students’ reasoning. If appropriate, you could also use this problem to encourage students to represent the concepts algebraically and set up equations.

**Possible approach**

This problem should be used once students have already seen the mean, median mode and range. As a starter, you could give students some sets of numbers and ask them to find the mean, median, mode and range. Then, ask them to find three numbers with mean 3 and mode 2. Is there more than one way of doing it?

After the first example, invite students to share their strategies.

Students could then work in pairs or small groups to find numbers for each set of conditions. You could encourage them to record their strategies, and whether they used the same strategy for all of the questions. At the end of the activity, students could share their strategies and conclusions in a whole class discussion.

Note that number 5 is only possible if you allow 0 as one of the numbers, and in that case there are several possible solutions. There are two possible solutios for number 6 if 0 is allowed.

Manipulatives could be used to introduce this work. Number cards could be used to allow students to visually see the problem.

**Key questions**

What is the difference between the averages i.e. How is the mode different to the mean?

Is there only one answer?

Is there a method that always works?

Of the mean, median, mode and range, which should you focus on first as you choose your numbers, and why?

**Possible support**

You could begin by giving students some of the numbers, for example 2, ___, 5 (choose the third number so that the mean is 3 and the mode is 2).

Alternatively, you could show them helpful examples: you could give them some sets of three numbers with mode equal to 2, such as 2, 2, 3 and 1, 2, 2 and ask them to find the mean and the mode, before asking them to find three numbers with mode 2 and mean 3.**Possible extension**

Ask students to create their own conditions that can be presented to their peers.

Ask students to investigate when it is possible to find numbers that meet their conditions, and when it is impossible. Can they create impossible conditions that look possible?