Belarusian Mathematical Olympiad

a. See Problem B.4.

b. Let it be possible to choose six four-digit numbers satisfying the problem condition. By the above, any of digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ must be exactly in three of these six numbers.

Further, it is not hard to prove that no two of the chosen numbers have the same three digits.

Let $a_1,a_2,a_3,a_4,a_5,a_6$ be the chosen six numbers. Let $P_i$ denote the set consisting of six pairs formed by the digits of $a_i$, $i=1,\dots ,6$. It is easy to see that the set ${\bigcup }_{i=1}^6{P}_i$ has exactly $28$ elements. Moreover, either $P_i\cap P_k=\varnothing$ or $P_i\cap P_k$ has exactly one element, and $P_i\cap P_j\cap P_k=\varnothing$, for all distinct $i,k,j$.

By inclusion-exclusion formula ($N(X)$ is the number of elements of the set $X$) $$N\left(\bigcup _{i=1}^6{P}_i\right)=\sum _{i=1}^{6}N(P_i)-\sum _{\begin{array}{c}i,j=1\\ i<j\end{array}}^{6}N(P_i\cap P_j).(\ast )$$ Since $N\left({\bigcup }_{i=1}^6{P}_i\right)=28$ and $N(P_i)=6$ for any $i=1,\dots ,6$, equality $(\ast )$ can be written as $$\sum _{\begin{array}{c}i,j=1\\ i<j\end{array}}^{6}N(P_i\cap P_j)=8.(\ast \ast )$$ Let $x$ be the number of pairs $(P_i,P_j)$, $i<j$, for which $N(P_i\cap P_j)=0$, then for the remaining $15-x$ pairs we have $N(P_i\cap P_j)=1$. Thus $(\ast )$ can be rewritten as $15-x=8$, or $x=7$.

It is not very hard to prove that $x\ne 7$. Thus $x\le 6$, and so $N\ge 7$.