Many of you have managed to work out some of the different sets of positive numbers which statisfy the three conditions: mean = 4, median = 3, and mode = 3.

Congratulation to Grace and Lydia from Hethersett High School who successfully found all the possible sets and convinced us that they had found them all with their excellent demonstration. Here is their answer:

The problem tells us that there are 5 positive whole numbers that have a mean of 4, a median of 3, and a mode of 3. What could these numbers be?

First of all, because the median is 3, we know that if we write the numbers in numerical order, the middle number has to be 3.

Because the mode is 3, there must be at least one other 3, and since we are writing this in numerical order, a 3 must at least occupy a place on one side of the middle 3.

1st number: Has to be equal to, or less than 3

2nd number: Has to be equal to, or less than 3

3rd number: Has to be 3

4th number: Has to be equal to, or greater than 3

5th number: Has to be greater than 3 (if the largest number was 3, the mean could not be 4)

Since the mean is 4, we know that all the numbers must add up to 20 (20 divided by the number of numbers (5) is 4). We can guarantee two 3s in the group, so the remaining numbers must total 14.

Now we need to make a list of all the possible groups whose starting number is less than 3, and (apart from 3) no numbers are repeated twice. The only time that this repetition would work, would be with three 3s, and two of another number, but no whole number multiplied by 2 equals 11 (20 minus three 3s).

Here is our list:

1, 2, 11

1, 3, 10

1, 4, 9

1, 5, 8

1, 6, 7 (1, 7, 6 = 1, 6, 7 because ultimately the numbers will be written numerically).

2, 3, 9

2, 4, 8

2, 5, 7

3, 3, 8

3, 4, 7

3, 5, 6

Written with the other numbers, the full sets are:

1, 2, 3, 3, 11

1, 3, 3, 3, 10

1, 3, 3, 4, 9

1, 3, 3, 5, 8

1, 3, 3, 6, 7

2, 3, 3, 3, 9

2, 3, 3, 4, 8

2, 3, 3, 5, 7

3, 3, 3, 3, 8

3, 3, 3, 4, 7

3, 3, 3, 5, 6

There are eleven sets. We have found all possible sets because we wrote them in ascending order, with the smallest possible number in first position, then the smallest possible in second position and the remainder in the last position. After that we continue to write the smallest number in first position till we had exhausted all possibilities for the other two numbers, then moved on to the
second smallest number in first position, and so on. Finally we deleted all repeating combinations, and all the final sets satisfy all conditions.

Jacob (Brewood Middle School) also found the possible set. But he found 17 possibilities in total because he included 0 in his answer. Some of you also made the same mistake of including 0 as a positive number.

0,1,3,3,13

0,2,3,3,12

0,3,3,3,11

0,3,3,4,10

0,3,3,5,9

0,3,3,6,8

1,2,3,3,11

1,3,3,3,10

1,3,3,4,9

1,3,3,5,8

1,3,3,6,7

2,3,3,3,9

2,3,3,4,8

2,3,3,5,7

3,3,3,3,8

3,3,3,4,7

3,3,3,5,6

When the range of 10 was added to the three original conditions, Ha-young (Wesley College), Chris and Matt (Staunton and Corse), Bethaney and Kieran (Staunton and Corse), James (Longthorpe Primary School), and Grace and Lydia (Hethersett High School) all sent in the correct solution.

Grace and Lydia followed their answer given above and concluded that:

If the range is 10 the numbers must be 1, 2, 3, 3, 11, because this is the only group with a range of 10.

Other good reasoning was sent by James (Longthorpe Primary School):

There must be five numbers so:

[?] [?] [?] [?] [?]

The median is three so:

[?] [?] [3] [?] [?]

The mode is also three so there must be at least two 3's so:

[?] [3] [3] [?] [?]

The mean is four so because there are five numbers, all the numbers, when added together, must make 20:

[?]+[3]+[3]+[?]+[?]= 20

The range must be 10:

First let's try 3 - 13;

[3] [?] [?] [?] [13]

This will not work. Because 3+13=16, we need 4 more to add up to 20.

We need two 3's, which makes 19.

This leaves only 1 and 2 empty spaces, which means 0.5 per space.

This cannot happen because each space has to be a whole number.

Now let's try 1 - 11.

There must be two 3's so, 1+11+6=18 or [1] [?] [3] [3] [11].

There is 2more to add up to 20 so

[1] + [2] + [3] + [3] + [11] = 20.

With the new set of conditions: mean = 31, median = 33, mode = 34, and range = 8, all of you, who have attempted to solve the problem have found the right answer. Congratulation to Daniel (Haberdashers' Aske's Boys' School), Rachel (West Moors Middle), Ha-young (Wesley College), Chris and Matt (Staunton and Corse), Bethaney and Kieran (Staunton and Corse), James (Longthorpe Primary School), and Grace and Lydia (Hethersett High School). This is a very clear explanation from Grace and Lydia:

The second question tells us that another set of 5 positive whole numbers have a mean of 31, a median of 33, a mode of 34, and a range of 8.

Once again, the middle number has to be 33, and both places numerically above it must contain a 34, since at least two 34s are needed, and there are only 2 spaces a 34 could go into without affecting the median.

1st number: Has to be less than 33.

2nd number: Has to be less than 33.

3rd number: Has to be 33.

4th number: Has to be 34.

5th number: Has to be 34.

Because the largest number is now 34, and the range is 8, the smallest number has to be 26 (34 - 8 = 26).

To find the final number, as the mean is 31, we must multiply 31 by 5 to get the total of all 5 numbers. If we add up the 4 numbers we have found, and then subtract this sum from 31 x 5, we will get the final number:

155 - 127 = 28.

So our set is 26, 28, 33, 34, 34

This is the only set of numbers there is because the largest three numbers have to be where they are to satisfy specifications, and this fixes the first number through the range. This leaves only one number to be decided, and if that number is not 28, it will not satisfy the mean of 31. Therefore only one set of numbers meets the requirements.

Jacob also tackled the problem with the same method but different presentation:

There must be five numbers so:

[?] [?] [?] [?] [?]

The median is 33 so:

[?] [?] [33] [?] [?]

The mode is also 34, so there must be at least two 34's, so:

[?] [?] [33] [34] [34]

The mean is 31 so because there are five numbers, all the numbers, when added together, must make 155 (31x5):

[?]+[?]+[33]+[34]+[34]= 155

Now the numbers must add up to 155 and there also must be two 34's. Also, the range must be eight and our highest number is 34 so, 34-8=26, so our lowest number must be 26:

[26] [?] [33] [34] [34]

All the numbers we have so far add up to 127.

Because155-127=28, 28 must be our final number.

[26] [28] [33] [34] [34]

There must only be one answer to this when there is a range as well.

Ha-young had a convincing reason why that is the only set we can find:

I am convinced that the set 26, 28, 33, 34, 34 is the only set we can find because 33 needs to stay in the middle and at least two 34 needs to stay, so we cannot have higher number. If we do, the median will change. The range also needs to be 8. Therefore the minimum number has to be 26.

Students from Kellett School in Hong Kong had a go at the **Final Challenge**:

Sean found the numbers 1, 2, 2, 2, 3 and 2, 5, 5, 6, 7 and 4, 8, 8, 8, 12 fit the criteria.

Sebastian found the numbers 2, 4, 4, 4, 6 and 3, 6, 6, 6, 9 fit the criteria.

Oliver found that 3, 4, 5, 5, 8 works too.

Students from Woodcot Primary School in Gosport found almost all the solutions to the Final Challenge:

1, 2, 2, 2, 3

2, 4, 4, 4, 6

3, 4, 5, 5, 8

2, 5, 5, 6, 7

3, 6, 6, 6, 9

3, 7, 7, 8, 10

3, 8, 8, 10, 11

4, 8, 8, 8, 12

5, 6, 8, 8, 13

5, 8, 9, 9, 14

George Simpson, from St. Gabriel's High School in Bury, found one more solution:

4, 9, 9, 10, 13

Mrs O’Clee class at Berkhamsted School completed the set with:

4, 6, 7, 7, 11

Well done to you all.