1. mean ~~~~ Add up all the numbers and divide by the number of numbers.
2. median ~~ Put the numbers in numerical order and pick the middle number.
3. mode ~~~ Find the numbers that appear most frequently.

The Revenge of the DAR puzzle describes an unusual case wherein the mean is twice the median (while making the mode moot, since all the test scores are different from one another, each appearing only once). 
 
 

We might reasonably suppose that the blue-book grades are percentages.  Better still, let's make that an explicit assumption and, while we're at it, that scores are all integers.  Now, if the number of students is odd, the median will be one of the scores and therefore an integer.  If the number of students is even, it is customary to average the two middle-most numbers to calculate the median.  That might result in a non-integer median.  We shall put this possibility to the side for the moment.

ure enough, the instructor sees that the grade average is raised by the student who is described by classmates, without affection, as a DAR.  We note that a DAR, being the outlier, has no effect on the median.  To have such a pronounced effect on the mean, the DAR's score must be as high as possible.  That would be no more than 100% by our assumption.  Meanwhile, all the other scores on the blue-book test must be low enough, so that taken together they will not elevate the median and prevent the DAR from having the effect on the mean specified in the Revenge of the DAR. 

The lowest possible grade, of course, is zero, and only one student can have it.  Whatever the lowest grade in the class, though, the rest of the grades must be integral amounts higher, making each score unique.  The smallest separation, of course, is one.  If the number of students is even, a median formed by averaging the two middle-most numbers will not be an integer.  So we conclude that the number of students is odd.  Computed as exactly twice the median, the mean is likewise an integer.

Let n = the number of students, and s[i] be the score of the i-th student. 
From what is known, we write...

mean = {s[1] + s[2] + ... + s[n]} / n 
median = s[(n + 1) / 2]
{s[1] + s[2] + ... + s[n]} / n = 2 s[(n + 1) / 2]
s[n] = s[DAR]
s[DAR] = 2 n s[(n + 1) / 2] - {s[1] + s[2] + ... + s[n-1]}

The instructor must have been disappointed with those students.  Except for one.
 

n =	3min	3max	5min	5max	7min	7max	9min	9max	11min	11max	13
lowest	0	23	0	14	0	9	0	5	0	2	0
median	1	24	2	16	3	13	4	9	5	7	6
mean	2	48	4	32	6	26	8	18	10	14	12
DAR	5	97	14	98	27	99	44	94	65	89	90

The table above shows the range of possible scores for n < 13.  For n > 13, the conditions of the Revenge of the DAR puzzle cannot be met.  If n = 15, say, the DAR would have to be given bonus points to reach 119%.

As for the 'revenge' part, suppose that a passing grade requires a score of at least half the class average.  How many of the 13 students flunk the exam as a consequence of the DAR's aggressive study habits?