Solutions:

1. They must all be 20. Any of the other statistics would reveal this fact for sure.

2. The mean would be enough to determine how many 20s and how many 21s there are. For example, if there are four 20s and three 21s, then the mean would be about 20.4. Knowing the median would not be enough because a median of 20 for example could result in a situation in which there are three 20s and four 21s or five 20s and two 21s.

3. Suppose there are three numbers and two distinct ones. If you know that the minimum is 20 and the median is 23, then you know that the other number is also 23. On the other hand, if you know that the mean is 22, this could result from 20, 20 and 26.

4. The minimum is 20, the maximum is 22. Mean is 21.2. This could be explained by three 22s and two 20s or by two 22s, two 21s and one 20. The median will tell you which.

5. The point to which the total distance is guaranteed to be at a minimum is the median. If there are an odd number n of values altogether, then this is the absolute minimum.

Here is why. Suppose the median is m. Let the total distance to m be x. Now consider the total distance to m+1. Because m is a median, there are (n-1)/2 values that are m or less. So, at least 1+(n-1)/2 values increase their distance by 1 when calculating the total distance to m+1. At most (n-1)/2 values decrease their distance by at most 1 when calculating the total distance to m+1.

So if there are 17 numbers and the median is 35, the mean is 34 and the distance to 38 is x+5, then there is one instance of 35, no instances of 36, and one instance of 37.

Here is why: if there were one instance of 35 and zero instances of 36 or 37, then the distance to 38 would increase by 3 * 8 = 24 for the numbers below 35 and decrease by 3 * 8 for the numbers greater than or equal to 38. The distance for 35 itself would increase by 3. This would give a net increase of 3. The fact that there is a net increase of 4 must mean that there is an instance of 37. In this case the net decrease for the upper eight values would be 3 * 7 = 21 for those greater than or equal to 38 and 1 for 37. This implies that there are seven values that are 38 or greater, one 35 and one 37. Because the mean is 34, the total distance of the numbers greater than 34 is at least 1 + 3 + 28, for the 35, the 37 and the seven values that are 38 or greater. This total is 32. The values below the mean must match this, so they must all be 30s. Since they cannot be less than 30, the uppermost value cannot be greater than 38. So there are eight 30s, one 35, one 37, and seven 38s.

6. The reasoning is exactly the same as for 17 numbers with one 35, one 37, 849 38s and 850 30s.