The answer is 6.Let the 8 numbers be a1 <= a2 <= a3 <= … <= a8 (NOTE that we don’t even need to employ the restriction to ‘WHOLE numbers’ - we can prove this in full generality over the real numbers).Then we are given that a1 = 2, and we WISH to maximise a8, SUBJECT to the constraints on the mean, median and mode.Since the median is 3, we have a4 + a5 = 6. But a5 >= a4, so a5 >= 3 (since adding a5 to both sides gives us that 2a5 >= a4 + a5 = 6).Though, if a5 > 3, we would have that a4 = 6 - a5 < 3, and then it would be impossible for the mode to be 3 (since none of the numbers could be 3).Hence, a5 = 3.This, in turn, FORCES a4 to be 3.Now, we have that a2 and a3 are at least 2, so a2 + a3 >= 4.Moreover, we have that a6 and a7 are at least 3, so a6 + a7 >= 6.Let S = a1 + a2 + a3 + … + a8.Then, we have that S = 24, since the mean is 3, and the mean is S divided by 8 (so 3 = S/8 implies that S = 3•8 = 24).Then S = 2 + a2 + a3 + 3 + 3 + a6 + a7 + a8 >= 2 + 4 + 6 + 6 + a8.So S = 24 >= 18 + a8, so a8 <= 6.So perhaps 6 is the maximum possible value of a8.If we try the dataset 2, 2, 2, 3, 3, 3, 3, 6, we have a mode of 3, a sum of 6 + 12 + 6 = 24 (and hence a mean of 3), and a median of 3.So, since a8 <= 6, and the fact that we have found a valid dataset with a8 = 6, we have maximised a8 and thus found the solutionStep-by-step explanation: