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Solution to M,M and M problem (nrich, June 2024)
Problem : find all sets of 5 positive integers such that:
Mean = 4, Mode = 3 and Median = 3
Solution must demonstrate how you know that ALL such sets have been found.
Solution :
Mode = 3
This implies that each set has a minimum of two 3’s.
Median = 3
This implies that the middle element of each set must be 3 (when written with elements in ascending order) as each set contains five integers.
Mean = 4
This implies that the element sum of each set is 20 from the definition of the mean.
Combining these three observations and applying the format of writing the set elements in ascending numerical order, we have three possible types of set:
Type 1 – two 3’s
(_, 3, 3, _, _ ) and (_, _, 3, 3, _)
In the first case, the leading number may be either a 1 or a 2. The remaining pair of numbers must be such that the sum of the set is 20, providing the following possible sets:
(1,3,3,4,9) (1,3,3,5,8) (1,3,3,6,7)
(2,3,3,4,8) (2,3,3,5,7)
Notice that (2,3,3,6,6) would violate the Mode = 3 condition.
The second case has one possible combination of starting values:
(1,2,3,3,11)
Again, notice that (1,1,3,3,12) and (2,2,3,3,10) would violate the Mode = 3 condition.
Type 2 – three 3’s
(3, 3, 3, _, _) or (_, 3, 3, 3, _) or (_, _, 3, 3, 3)
The first case provides the following sets:
(3,3,3,4,7) (3,3,3,5,6)
The second case of this type of set may again only start with a 1 or a 2, providing two more sets:
(1,3,3,3,10) (2,3,3,3,9)
The third case can only start with 1,2 without violating the Mode = 3 condition. This does not produce a viable set as the element sum would be less than 20 which would break the Mean = 4 condition.
Type 3 – four 3’s
(3, 3, 3, 3, _) and (_, 3, 3, 3, 3)
Only the first template provides a viable set without violating the chosen formatting of ascending order of elements and the sum of the set being equal to 20:
(3,3,3,3,8)
We do not have to consider the set consisting of five 3’s as this would violate the Mean = 4 condition.
By considering all possible types of set available as we increase the number of 3’s, we have provided a proof by exhaustion that our solution contains all possible sets:
(1,3,3,4,9) (1,3,3,5,8) (1,3,3,6,7) (2,3,3,4,8) (2,3,3,5,7) (1,2,3,3,11)
(3,3,3,4,7) (3,3,3,5,6) (1,3,3,3,10) (2,3,3,3,9) (3,3,3,3,8)