You may wish to explore Equal Averages before tackling this problem.
Can you find sets of five positive whole numbers that satisfy the following properties?
Mode < Median < Mean
Mode < Mean < Median
Mean < Mode < Median
Mean < Median < Mode
Median < Mode < Mean
Median < Mean < Mode
Not all of them can be done! Can you explain why?
Would it help if you could use six numbers instead of just five?
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Unequal Averages Solution
The question says you can input any five positive integers and that the output must be a single digit number. This means that the output must also be positive and lie in the range 19.
$\text{sum of quantities} = $\text{mean} \times $\text{number of quantities}
The first thing I did was to calculate the sum of quantities by multiplying the possible means (19) by the number of quantities, which is fixed at 5. I then used spaces to represent the five positive integers:
$n$  Possibly $x$  $x$  Possibly $x$  $n+x$ 
$A$  $B$  $C$  $D$  $E$ 
Let’s say that the mean is $x$. Since the mean must equal the median, then $x = \text{median}$, which must occupy position C. But as the median = mode, then $x = \text{mode}$, meaning that $x$ must appear at least twice, so $x$ can appear in B and/or D. If n occupies position A, then n + x will occupy position E, as $x =$ range which also equals the mode.
$A + B + C + D + E = 5x$ because $5x \div 5 = x$, where $x$ is the mean. Then $5x – 2x = 3x$
The mode will appear at least twice, so the remaining three numbers will sum to 3 times the mean.
Next I reduced the number of variables to just one space, which became the balancing figure. Here is how I did it:
(i) input a value x for the median into position C
(ii) add the same number to represent the mode (this could occupy B or D)
(iii) calculate a target sum of quantities using a mean in the range 1 – 9, e.g. $5 \times 9 = 45$
(iv) subtract the mode from the sum of quantities
(v) input a value into position A (start with 1 then 2 etc..), add this to the number x you put in C and use this to populate position E.
(vi) now only one space is left blank – this is the balancing figure (which could be in B or D)
(vii) test the model against the target sum of quantities (e.g. $5 x 9 = 45$, pretty quickly you will find all the available sets (including bimodal sets, which do not technically fulfil the criteria. These usually sit either side of any sets that do actually work).
Mean = Median = Mode = Range  Sets and Observations 
=1  none possible (can only input 1, but range = 0) 
=2  1,2,2,2,3 works 
=3  1,3,3,4,4 and 2,2,3,3,5 are bimodal (mode has two values). 
=4  2,4,4,4,6 works 
=5  2,5,5,6,7 and 3,4,5,5,8 work 
=6  3,6,6,6,9 works but 2,6,6,8,8 and 4,4,6,6,10 are bimodal 
=7  3,7,7,8,10 and 4,6,7,7,11 work 
=8  3,8,8,10,11 and 4,8,8,8,12 and 5,6,8,8,13 work 
=9  4,9,9,10,13 and 5,8,9,9,14 work, but 3,9,9,12,12 and 6,6,9,9,15 are bimodal 
Part Two: we still have five spaces labelled AE, but we now need to find values that will satisfy an inequality
Mode<Median<Mean
I started with the mean. 3 is the smallest mean possible, therefore the median would have to be 2 and the mode 1. This time I put 2 in space C, and 1’s in A and B. Then I put a 3 in D (median plus 1), leaving E as the balancing figure required to obtain an overall mean of 3. Once I found a set that worked, I increased D and reduced E to obtain further sets. I then worked systematically to substitute ever larger numbers for the mean. Here is a sample of what I found:
Mode<Median<Mean – Sets using 5 figures 

1<2<3 
1<2<4 
1<2<5 
1<2<6 
1<2<7 
1<2<8 
1<2<9 
1,1,2,3,8 
1,1,2,3,13 
1,1,2,3,18 
1,1,2,3,23 
1,1,2,3,28 
1,1,2,3,33 
1,1,2,3,38 
1,1,2,4,7 
1,1,2,4,12 
1,1,2,4,17 
1,1,2,4,22 
1,1,2,4,27 
1,1,2,4,32 
1,1,2,4,37 
1,1,2,5,6 
1,1,2,5,11 
1,1,2,5,16 
1,1,2,5,21 
1,1,2,5,26 
1,1,2,5,31 
1,1,2,5,36 

1,1,2,6,10 
1,1,2,6,15 
1,1,2,6,20 
1,1,2,6,25 
1,1,2,6,30 
1,1,2,6,35 

1,1,2,7,9 
1,1,2,7,14 
1,1,2,7,19 
1,1,2,7,24 
1,1,2,7,29 
1,1,2,7,34 


1,1,2,8,13 
1,1,2,8,18 
1,1,2,8,23 
1,1,2,8,28 
1,1,2,8,33 


1,1,2,9,12 
1,1,2,9,17 
1,1,2,9,22 
1,1,2,9,27 
1,1,2,9,32 


1,1,2,10,11 
1,1,2,10,16 
1,1,2,10,21 
1,1,2,10,26 
1,1,2,10,31 



1,1,2,11,15 
1,1,2,11,20 
1,1,2,11,25 
1,1,2,11,30 



1,1,2,12,14 
1,1,2,12,19 
1,1,2,12,24 
1,1,2,12,29 




1,1,2,13,18 
1,1,2,13,23 
1,1,2,13,28 




1,1,2,14,17 
1,1,2,14,22 
1,1,2,14,27 




1,1,2,15,16 
1,1,2,15,21 
1,1,2,15,26 





1,1,2,16,20 
1,1,2,16,25 





1,1,2,17,19 
1,1,2,17,24 






1,1,2,18,23 
The pattern leaps out: if the first four numbers are fixed, just increase the number in space E by 5 each time, so $(1,1,2,3,8)$ satisfies $1<2<3$ and $(1,1,2,3,13)$ satisfies $1<2<4$. This is because $P \div 5 = Q$, whereas $(P+5) \div 5 = (Q+1)$. And you can go on to find many sets: e.g. $(1,1,3,4,11)$, $(1,1,3,5,10)$, $(1,1,3,6,9)$ and $(1,1,3,7,8)$ satisfy $1<3<4$, and $(1,1,3,4,16)$, $(1,1,3,5,15)$, $(1,1,3,6,14)$, $(1,1,3,7,13)$, $(1,1,3,8,12)$, $(1,1,3,9,11)$ satisfy $1<3<5$ etc...
For four numbers I started with $1<2<3$. Put 1 in space A and B, and 3 in space C to make a median of 2. Mean of 3 leads to a sum of quantities of 12 ($3 \times 4 = 12$), balancing figure in D is 7. So $(1,1,3,7)$ satisfies $1<2<3$, $(6,6,8,12)$ satisfies $6<7<8$, $(5,5,9,17)$ satisfies $5<7<9$ etc...
Six numbers leads to six spaces labelled AF. Test $1<2<3$: put 1 in spaces A, B and C, then 3 in space D to generate a median of 2. I then chose to put 3 in space E, although you could also put in 4,5,6 in this case, leaving F as the balancing figure required to make an overall mean of 3. So $(1,1,1,3,3,9)$, $(1,1,1,3,4,8)$, $(1,1,1,3,5,7)$ and $(1,1,1,3,6,6)$ satisfy $1<2<3$. There are many possibilities if you use other variables.
Mode<Mean<Median
$G(G1)$  $G  (G1)$  $2G1$  $N$ 
$A$  $B$  $C$  $D$ 
Mean<Mode<Median
Mean<Median<Mode
This time the mode will go into spaces D and E, and the median into space C. The sum of numbers in spaces A and B become the balancing figure to achieve an overall mean. Five numbers work: e.g. $(7,12,13,14,14)$, $(8,11,13,14,14)$ and $(9,10,13,14,14)$ all satisfy $12<13<14$ which can also generate a four numbered set: $(8,12,14,14)$ and a six numbered set $(7,11,12,14,14,14)$, which is really cool!
Median<Mode<Mean
$G2$  $G1$  $G$  $G+1$  $G+1$ 
$A$  $B$  $C$  $D$  $E$ 
Median<Mean<Mode
$G2$  $G1$  $G$  $G+4$  $G+4$ 
$A$  $B$  $C$  $D$  $E$ 
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unequal averages answers
1) Mode < Median < Mean 1,1,2,3,5
2) Mode < Mean < Median 1,1,7,9,10
3) Mean < Mode < Median impossible
4) Mean < Median < Mode 1,2,9,10,10
5) Median < Mode < Mean impossible
6) Median < Mean < Mode 1,2,3,7,7
These are just the answers I got and I'm not saying these are the only answers to the puzzle. I'm not certain 3 and 5 are impossible but my findings show that they are.
prove it:
1) 1,1,2,3,5 mode: most often 1, median: in the middle 2, mode: add them up and divide by the number of numbers 2.4
2) 1,1,7,9,10 mode: most often 1, mean add them up and divide by the number of numbers 5.6, median middle number 7
4) 1,2,9,10,10 mean: add them up and divide by the number of numbers 6.4, median: middle number 9, mode: most often 10
6) 1,2,3,7,7 median: middle number 3, mean: add them up and divide by the number of numbers 4, mode: most often 7
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Working it out
To show you how I worked this out I will use number 1 as an example. (Mode < Median < Mean) As the smallest must be the Mode (the one that appears most often), I chose to work this out first. I chose number 2. The next one is the median. For this one you have to choose all five numbers. I chose 2, 2, 3, 4, 5. This made the median 3. Finally I worked out the mean of these numbers. This was 3.2. 2<3<3.2. This would also work with 1,1,2,3,4 or ,if you were using six numbers, 1,1,2,3,4,5 1<2.5< 2.666666666666667 could work.
For number 2 (Mode < Mean < Median) I would suggest working out the mode, then the mean, then the median.
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