In India, there is an exam **JEE-Mains** which is taken by more than a million students. Here, there are 90 objective problems, each having 4 choices, out of which only 1 is correct. Each correct answer carries 4 marks, and each incorrect answer costs -1 marks. You get 0 marks for an unattempted problem.
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Consider a very special student sitting in the exam. This student knows at least 1 of the problems correctly. The probability that he knows the correct answers to exactly \(k\) problems is directly proportional to \(\frac{1}{k}\). In each of the problems he doesn't know the answer to, there is an 80% chance that he marks it randomly and a 20% chance that he leaves it unanswered.

Find the greatest integer less than or equal to the expected marks of this student.

**Details and assumptions**

\(\displaystyle \sum_{n=1}^{90} \frac{1}{n} = \frac{1}{0.196751}.\)

If he answers a problem randomly, the probability that he gets it correct is exactly \(\frac{1}{4}.\)

As any other student, if he knows the correct answer to a problem, he will mark it, i.e. he won't leave it unanswered.

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