6. In how many ways can the number 1024 be factored into three natural factors such that the first factor is divisible by the second, and the second is divisible by the third?
Answer: 14

Solution: Note that the factors have the form $2^{a} \times 2^{b} \times 2^{c}$, where $\mathrm{a}+\mathrm{b}+\mathrm{c}=10$ and $a \geq b \geq c$. Obviously, c is less than 4, otherwise the sum would be greater than 10. Let's consider the cases:

$c=0)$ Then $b=0, \ldots, 5, a=10-b-6$ options

c=1) Then $b=1, . .4, a=9-b-4$ options

$c=2) b=2,3,4, a=8-b-3$ options.

$c=3) b=3, a=4-1$ option.

In total $6+4+3+1=14$ options.