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First, consider the possible numbers for A, B, and C. Naturally, all three are positive integers, and A≥B≥C. Also, A is not equal to B, or the teacher would not be able to pick two combinations of B questions.
Before hearing what other students say, I would need to study all but B-C questions (the number of questions on the test that I may ignore). This is A-B+C. Note that this is not true if C=0 (but C can't equal 0).
After hearing the news from the other students, the number of questions I must study is calculated differently. The teacher must give me at least one of the questions he did not give to the other students. It would be best for me to study all these questions (A-B).
B questions have already been given to the other students. In the worst case scenario, B-1 of these used questions will be given to me. I will have to answer at most C-1 of these questions. According to the A-B+C equation shown earlier, I will have to study (B)-(B-1)+(C-1), or C of these questions. However, if C=1, I won't have to study any of them.
If C is not 1, then the total number of problems that need studying is A-B+C, the same as before. If C is 1, then it is A-B, which is C less than before, or rather one less than before.
Therefore, N=0 for all cases, except when C=1, where N=1. |
