Imagine the seven integers are placed in ascending order. The fourth must be a 7 (because the median is 7) and there must be at least two 5s.

Therefore the list must be either

(A) 5 – 5 – X – 7 – X – X – X or (B) X – 5 – 5 – 7 – X – X – X

If (A), then the list must be either 5 – 5 – 5 – 7 – X – X – X or 5 – 5 – 6 – 7 – X – X – X

We also know that the total of the seven integers must be a multiple of 7, in order to have an integer mean.

For 5 – 5 – 5 – 7 – X – X – X , the least range is when the list is completed as 5 – 5 – 5 – 7 – 7 – 8 – 8, but this has a total of 45, which is not a multiple of 7. The lowest possible multiple of 7 is therefore 49. This total can be only be achieved by completing the list with 10 as the largest number, as if 9 is the largest number the total can only reach 48 (5 – 5 – 5 – 7 – 8 – 9 – 9)

So (for example) 5 – 5 – 5 – 7 – 7 – 10 – 10 would work, and has a range of 5.

For 5 – 5 – 6 – 7 – X – X – X , the least range is when the list is completed as 5 – 5 – 6 – 7 – 8 – 9 – 10, which has a total of 50, meaning that the lowest total we can consider is 56. Clearly this can only be done if the range increases, so there is no smaller range to be found here.

If (B), then the list has to be X – 5 – 5 – 7 – 8 – 9 – 10, where the largest possible value of X is 4. This would always give a range larger than 5, no matter how the first value is adjusted, so there is no smaller range here.

Therefore, the smallest possible range is 5.

Challenge 322: Range Riddle