Source: *Mathematics Magazine*, 1958, pp. 237.

**Hint:** A Wonder: What is so special about 26?

Get your feet wet with the problem by systematically trying cases for a N-problem test, where N = 1, 2, 3, .... For example, for a 4-question test, your cases are WWWW (-20), WWWC (-7), WWCC (+6), WCCC (+19), and CCCC (+32), where order does not matter.

Focus on connecting the CCC...C (for n C's) option in each row with the WCCC...C (1 W and n C's). You should see diagonal patterns appear that can lead to a solution.

Or, if this hint makes no sense, there are multiple other ways to approach the problem...

**Solution Commentary:** First, for the pattern suggested in the hint, one would need the original number of C's to be 5, giving a score of +40. Then as you add another wrong W (or -5) to this in each subsequent row, you will eventually get a score of 0. This will take 8 rows beyond the original row 5, making a total of 13 questions. This pattern will repeat for 26 questions, etc...why?

As another solution approach, for N questions, you could let W and C represent the number of answers wrong and correct respectively. Then, W+C = N and we are interested in the case -5W+8C = 0. Solving this system, we find W = (8/13)N. When N = 26, there are 16 wrong answers and 10 correct answers.

Or, a clever solution is provided C.W. Trigg, the problem's author (and a great problem poser/solver). He writes: "The number of answers in each category is inversely proportional to the value, so there were (5/13)(26) or 10 correct answers." Stop and think about this...and it may start to make sense, especially if you remind yourself of the mathematics in the previous solutions.