For this problem, stop and think before rushing into possible solution attempts. Some calculations may be necessary, but think first...and after these calculations.
On a 26-question test, 5 points were deducted for each wrong answer and 8 points were credited for each correct answer. If all 26 questions were answered, how many correct did Stu Dent get if his score was zero?
Again, stop and think...a lot.
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.