18. 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.

In a trapezoid, the diagonals of which intersect at a right angle, it is known that the midline is 6.5 and one of the diagonals is 12. Find the second diagonal.

## Answer 5

Solution: Parallel translate one of the diagonals so that it forms a right triangle with the other. Then in this triangle, one leg is 12, and the hypotenuse is 13, so the remaining leg is 5.