Challenge Level

Most solutions we received included these ideas. This is Euan's work, from Long Field Academy in the UK:

Firstly, having a mean of 4 meant the numbers had to add up to 20 because there were five numbers, and 20$\div$5=4. Then with the mode being 3 it meant that there had to be more than one three and no duplicates of other numbers. With the median being 3 it meant that [3 had to be the middle number].

Harry from Hull Collegiate School, Yuna, Vivienne and Jemima from Bute House Prep School and Rebecca and Thomas from Long Field Academy, all in the UK, tried out all of the different possibilities for the first number. This is Yuna's work:

If the first number was a 1 then you could have 1, 3, 3, 3, 10. Then, working systematically, I added one to the penultimate number and subtracted one from the final number. Using this way, I know that the average will always be 4 and that the mode and median will stay the same as long as I keep at least two 3s.

1, 3, 3, 3, 10

1, 3, 3, 4, 9

1, 3, 3, 5, 8

1, 3, 3, 6, 7

Then, continuing with this method, I replaced the 1 at the beginning with a 2. I got these answers…

2, 3, 3, 3, 9

2, 3, 3, 4, 8

2, 3, 3, 5, 7

*2, 3, 3, 6, 6 is not included because it isn't clear whether the mode is 3 or 6.*

Once again, I replaced the first digit and made it into a 3. I got the answers...

3, 3, 3, 3, 8

3, 3, 3, 4, 7

3, 3, 3, 5, 6

Thomas chose the largest possible number as the first number and worked down. Thomas' first set is the only one that Yuna missed:

Working from 11 down to 1 as the starting number and always having 3 as the next two numbers, use number bonds facts to work out the final 2 numbers needed to make the whole total 20.

11, 3, 3, 1, 2

10, 3, 3, 1, 3

*and so on.*

Millie from Nascot Wood Junior School in England and Eden-Rose from Bute House Prep School tried out the different numbers of 3s. Eden-Rose wrote:

The median shows us that our 5 numbers will look a bit like either: ?, ?, 3, 3, ? or: ?, 3, 3, ?, ?

Let's start with the first way. The first numbers must be 1 and 2. This is because if the first numbers were 1 and 1 or 2 and 2 then the mode would have to change. So we [have to] do 1, 2, 3, 3, 11.

Let's try it the other way now: ?, 3, 3, ?, ?. We could do 1 or 2. Let's be systematic and go with 1.

1, 3, 3, ... The other two numbers need to add up to 13. We can use 4, 9 or 5, 8 or 6, 7.

Now let's try it with 2: 2, 3, 3, 4, 8 or 2, 3, 3, 5, 7.

Sanika from India used number bonds to 14:

First, I let the five numbers be $a, b, c, d$ and $e.$

Since the median was 3, $c$ =3.

It was also mentioned that the mode was 3, which meant that at least one other variable must equal 3. So, I let $a$ be 3.

The mean being 4, meant that $(a + b + c + d + e)\div$5 = 4. Substituting in the values of $a$ and $c$ I got 14 to be equal to $b + d + e$. So I drew a table to find out all the possibilities.

$b$ | $d$ | $e$ |

1 | 2 | 11 |

1 | 3 | 10 |

1 | 4 | 9 |

1 | 5 | 8 |

1 | 6 | 7 |

2 | 3 | 9 |

2 | 4 | 8 |

2 | 5 | 7 |

3 | 3 | 8 |

3 | 4 | 7 |

3 | 5 | 6 |

The above are all the possibilities; however, we can permute the contents of each row in the above table to give us 24 sets for each row (if all cells in that particular row contain a unique value).

For example, the first set [3 1 3 5 8] can be rewritten as [1 5 3 3 8]. $c$ alone will not change its value as 3 is the median.

Mathew from Montmorency Secondary College in Australia wrote a method in terms of algebra:

Ordered list: $a, b, 3, c, d.$ $\{a, b, c, d\}$ all positive integers.

Fact 1: $a\leq b\leq 3\leq c\leq d,$ to keep list ordered.

Fact 2: $a+b+c+d=17,$ to make the average $4,$ keeping $3$ in the middle.

Case 1:

If $a=b=c=3$

$d$ must be $8,$ see Fact 2.

Case 2:

If $b=c=3$

$a+d=11$ (so fix average at $4$), $a\lt3\lt d.$

Follow rule until you exhaust the possibilities.

Case 3:

If $b=3$

$a+c+d=14$

$a\lt3\lt c\lt d, c\neq d$ (as this would make the set bimodal)

Follow this rule until all possibilities are exhausted.

Case 4:

If $c=3$

$a+b+d=14$

$a\lt b\lt 3\lt d, a\neq b$ (as with Case 3)

*The only case that is missing is if $a = b= 3$*