Bariz - Problem

We give the following construction. Suppose we have $x$ cards of colour $a$, $y$ cards of colour $b$ and $z$ cards of colour $c$ Now label the squares in the following way (see attachment) Now put cards in the squares as follows: Colour $a$ in squares labelled $1$ to $x$, colour $b$ in sqaures $x+1$ to $x+y$ and colour $c$ in squares $x+y+1$ to $100$. However we do not do this in a random order. Note that as no two numbers $\le50$ share a side, and no two numbers $>50$ share a side, the only colour cards at risk of sharing a side are colour $b$. So label the colours such that colour $b$ has the fewest cards. We see that as the smallest difference between any two numbers in adjacent squares is $45$ and colour $b$ has at most $33$ cards, there will never be two adjacent cards with colour $b$.