31st Hellenic Mathematical Olympiad

Number $1$ can be colored in two ways, white or blank. Number $2$ also can be colored white or blank. Then all even numbers $2,4,6,8,10,12,14,16,18,20$ have to be colored with the color of $2$. Also all numbers having common divisor greater than $1$ with the above numbers must be colored with the color of $2$. The remaining numbers greater than $10$, that is, $11,13,17,19$ can be colored in two ways. Therefore the coloring can be made with $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^6=64$ different ways. However we must delete the two cases we color all numbers white or blank. So we have finally $62$ different ways.