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##### Age 11 to 14 Challenge Level:

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:
$0,4,9,11$

Galal, from the Continental School of Cairo in Egypt had:
$0,3,8,11$

Ashley, Thea, Charlie and Vikash from Norwich School found:
$0,6,7,11$

Class 2C from The London Oratory School also found this, as well as the following:
$0,5,8,11$

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

6. The mode of $3$ tells us that there are at least two $3$'s. The mean of $4$ means that the numbers all add up to $20$, so the other three add up to $14$. As the range is $9$, this means that the only possible numbers are $1$, $3$ and $10$.

Therefore the overall set is: $1,3,3,3,10$

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 .

For the extension problems, Nour from The Continental School of Cairo in Egypt, Zach, Class 2C from The London Oratory School and Max from Castle School all found the following solutions for four, five and six numbers:

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?