Explanation:
x1, x2, x3, x4, x5
(x1 + x2 + x3 + x4 + x5)/5 = 4
x5 - x1 = 6
x3 = 3
2 appears more often than any other number (at least twice), otherwise there would have been multiple modes.
x1 + x2 + x3 + x4 + x5 = 4×5 = 20
x3 = 3
x1 + x2 + 3 + x4 + x5 = 20
x1 + x2 + x4 + x5 = 17
x5 = x1 + 6
x1 + x2 + x4 + x1 + 6 = 17
2×x1 + x2 + x4 = 11
since we did not have the number 2 yet (and we made x5 the harvest number, which has to be larger than 2), and it has to appear at least twice.
so, two numbers out of x1, x2, x3 must be 2. not all three of them can be 2, because 2×2 + 2 + 2 = 8 and not 11 as the last equation demands.
we have therefore 2 different solutions, as either
x1 and x2, or x2 and x4 can be 2.
x1 and x4 is the same solution per the last equation as x1 and x2 (as the variables are interchangeable for the number assignment, only x1 is different, because it has an additional factor in the last equation).
solution 1 (x1 and x2 are 2) :
2×2 + 2 + x4 = 11
4 + 2 + x4 = 11
6 + x4 = 11
x4 = 5
so,
x1 = 2,
x2 = 2,
x3 = 3
x4 = 5
x5 = x1 + 6 = 2 + 6 = 8
solution 2 (x2 and x4 are 2) :
2×x1 + 2 + 2 = 11
2×x1 + 4 = 11
2×x1 = 7
x1 = 7/2
so,
x1 = 7/2
x2 = 2
x3 = 3
x4 = 2
x5 = x1 + 6 = 7/2 + 12/2 = 19/2