Macedonian Mathematical Competitions

Since $30^{2008} = 2^{2008} \cdot 3^{2008} \cdot 5^{2008}$ and $20^{2007} = 2^{4014} \cdot 5^{2007}$, all the numbers divisible by $30^{2008}$ and not divisible by $20^{2007}$ are:

1) $2^{k} \cdot 3^{l} \cdot 5^{m}$, $l = 1, 2, \ldots, 2008$, $k, m = 0, 1, 2, \ldots, 2008$ or $2008 \cdot 2009^{2}$ numbers,

2) $2^{k} \cdot 5^{m}$, $m = 2008$, $k = 0, 1, 2, \ldots, 2008$ or $1 \cdot 2009 = 2009$ numbers.

Finally, there are $2008 \cdot 2009^{2} + 2009$ numbers divisible by $30^{2008}$ and not divisible by $20^{2007}$.