Prove that at least one student solved all the problems using PHP

Question

In a physics exam, $5$ problems were given to a class of $N$ students. Suppose every two of these problems were solved by more than $\frac{3N}{5}$ students.Prove that at least one student solved all the problems

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I have used PHP here. We know if at least $(k.n +1)$ objects are distributed among $n$ boxes, then one of the boxes must contain at least $(k+1)$ objects.
So according to this problem $(k.n +1)$ = $2$ |where $n$ = $\frac{3N}{5}$.
But unable to prove at least one student solved all the problems.
I have to prove basically $(k+1)$ = $5$. It will be helpful for me if someone helps me with this.

Answer 1

Actually we don't need this:

Say student $S_i$ solved $0\leq d_i\leq 5$ problems and every problem $P_i$ vas solved by $p_i$ students. Then we have $$\sum_{i=1}^n d_i =\sum_{i=1}^5 p_i $$

But this is relevant:

On the other hand we have for each pair of problems $P_i,P_j$ that $p_{i,j}> 3n/5$ solved them both and every student $S_i$ solved ${d_i\choose 2}$ problems, so: $$ \sum_{i=1}^n {d_i\choose 2} =\sum_{i,j} p_{i,j} > 10\cdot {3n\over 5} = 6n $$ Suppose $d_i\leq 4$ for each $i$, then we have $$\sum_{i=1}^n {d_i\choose 2} \leq 6n$$ A contradiction.

New version of the same solution, less formal:

So we have $10$ pairs of questions, so there is in total more than $10\cdot {3n\over 5} = 6n $ correctly answered pairs of questions and suppose no one answered more than $4$ questions, so everyone answered at most ${4\choose 2}=6$ pairs of questions. But then in total they answered at most $6n$ pairs of questions. A contradiction.