Club of Abusers

First, we claim that $\boxed{4}$ exams is enough.

• For the first exam, answer everything True.
• For the second exam, answer everything False.
• For the third exam, answer everything True, except the last question which is answered False.
• For the fourth exam, answer everything False, except the last question which is answered True.

Let $c_1, c_2, c_3, c_4$ be the number of correct answers on the first, second, third, and fourth exams respectively.

Note that $c_1 + c_2 = 100$ ; each question is answered correctly by exactly one of first exam and second exam. If either of $c_1, c_2$ is less than $50$ , then the other will necessarily be greater than $50$ , so you win. Thus if you still don't win, $c_1 = c_2 = 50$ .

Next, note that $c_3 = c_1 \pm 1$ , since a difference of a single question will either make a correct answer incorrect, or an incorrect answer correct; the number of correct answer changes by exactly $1$ . If $c_3 = c_1 + 1$ , we have $c_3 = 51$ and we're done. Otherwise, $c_3 = c_1 - 1 = 49$ . But then $c_3 + c_4 = 100$ once again, so $c_4 = 51$ and we're done.

Next, we claim that $3$ exams is not enough.

Suppose there is such way to take only $3$ exams and guaranteeing a win. For each question, there are $3$ exams and $2$ choices; by Pigeonhole Principle, we can take one choice that is chosen at least twice. Assume that the correct answer is the other choice.

If all exams have no more than $50$ correct answers, then we're done. Otherwise there is such exam; without loss of generality, suppose that it is the first exam. Now choose any $50$ questions that the first exam answers correctly, and fix their answers; for the remaining questions, make their correct answers to be opposite of what the first exam answered. By construction, the first exam only scores $50$ ; in addition, for the questions where the first exam answers correctly, the second and third exams don't answer correctly, and thus each of the second and third exams can score at most $50$ as well (since they have $50$ wrong answers already). This proves that $3$ exams is not enough.

Kindly help me out here My steps:

Step 1. Mark Everything true Possibility 1: Marks greater than 50 We enter the club Possibility 2: Marks less than 50 Step2: Mark everything False we enter the club Possibility 3: Marks equal to 50 Step2: We mark Last question false Sub Possibility 1: If we get 51 i.e. last question was infact False We enter the club Sub Possibility 2: If we get 49 i.e. last question was infact True Step3: Mark First 99 Falso and last one True We Enter the club

I see 3 steps are enough In the solution given step 2 is redundant

Once we know how many are true, why are we again calculating how many are False ?

It doesnt matter how you will answer everything, as long as the two sets of questions are entirely opposite of each other, and the other two sets are the same for the 1st two sets except one certain question is changed for both.