A math contest consists of a Part-1 and a Part-2, with a combined total of 28 questions. Each contestant solved 7 problems altogether. For each pair of problems there were exactly two contestants who had solved both of them . Prove that there was a contestant who in part 1 solved either no problem or at least 4 problems .
In my attempt , I begun by trying to find out the number of contestants. Let us assume that some problem $A$ was solved by $x$ people. As each of these $x$ people will have solved 6 other questions, 6$x$ other problems were solved (With repetition).There are 27 problems other than $A$. As each of the others is solved twice, 54 other solutions exist, hence
$6x=54$ so $x=9$.
That would mean that each problem is solved by 9 people and there are 28 problems .Pair this with the fact that each person solves exactly 7 ,That leaves
$9×28/7 = 36$ contestants
It is not neccecsary that Part-1 and Part-2 have the same number of questions.
This is as far as I could get.Thank you
Your proof is correct. (There is another way by counting directly the number of resolutions of pairs of problems : if you have $N$ constestants, each solves $7$ problems, so in total $\binom{7}{2} N = 21 N$ pairs of problems were solved, and since each pair is also solved by exactly $2$ contestants, this has to be equal to $2\binom{28}{2} = 28 \cdot 27$ and therefore indeed $N = 4\cdot 9 = 36$. Your way gives, furthermore, the number of contestants who solved one given problem.)
To conclude, assume by contradiction that everyone solved between $1$ and $3$ problems in part 1. (I assume parts 1 and 2 have the same number of problems, $14$, but correct me if this is not a hypothesis.)
The $14$ problems of part 1 were resolved $14 \cdot 9$ times, by $36$ contestants. But $\frac{14 \cdot 9}{36} = \frac{7}{2} = 3.5$. This is impossible if everyone solved less than $4$ problems. (By the way, you prove that there always exists a contestant who solved at least $4$ problems ; the other option "there exists a contestant who solved no problem at all" does not matter.)
EDIT : How to tackle the fact that both part may have a different number of questions : Let $m$ be the number of problems in part 1. By doing the same as above, we can see that if $m > \frac{36}{9} 3 = 12$, then there is a contestant who did at least $4$ problems in part $1$.
Else, $m \leq 12$. Let $n_1, n_2, n_3$ be the number of contestants who did $1$, $2$ or $3$ problems in part 1, respectively. $n_1 + n_2 + n_3 = 36$. Then, by counting pairs of problems inside part 1, you see that you must have :
$$ n_2 + n_3 \binom{3}{2} = 2 \binom{m}{2} $$
Furthermore, $n_1 + 2 n_2 + 3 n_3 = 9m$ (each problem is solved by exactly $9$ contestants). Therefore, $n_2 + 2 n_3 = 9m - 36$ and then $n_3 = 2 \binom{m}{2} - 9m + 36 = m^2 - 10m + 36$. Then, you have :
$$ n_2 = 2 \binom{m}{2} - 3 n_3 = -2m^2 + 29m - 108 $$
And finally $n_1 = 36 - n_2 - n_3 = m^2 -19m + 108$.
These are polynomials of second order in $m$. But the polynomial for $n_2$ has $\Delta = 29^2 - 4 \cdot 108 \cdot 2 = -23$ so $n_2 < 0$ for every $m$. This is a contradiction.
(I hope I didn't make any mistake in my computations, please notify me if I did. Actually, I didn't need to distinguish between $m > 12$ and $m \leq 12$.)