Answer: $2^{6}-2=62$ ways.

Solution. Note that all even numbers must be of the same color. Since among them are the numbers 6, 10, and 14, the numbers divisible by 3, 5, and 7 must be of the same color. The remaining numbers are $1, 11, 13, 17$, and 19. Note that they can be distributed in any way among the two colors. Thus, we have 6 groups, each of which can be of any color, i.e., a total of $2^{6}$ ways of coloring.

Note that 2 of these do not fit, where all numbers are of the same color. In total, we get $2^{6}-2=62$ ways.

## Criteria

The most suitable criterion is used:

7 points. Any complete solution to the problem.

3 points. Noted that even numbers, as well as numbers divisible by 3, 5, and 7, must be of the same color.

1 point. The correct answer is given.

The following errors in an otherwise correct solution result in point deductions:

-1 point. Not considered that both colors must be present.

-2 points. Not considered that both colors must be present, and not considered that "1" is not a coprime number.

-2 points. Not considered the alternative coloring, i.e., there are two pairs of groups: 1 group - red, 2 group - blue and 1 group - blue, 2 group - red.

-1 point. When considering monochromatic groups, the number 1 or one of the prime numbers greater than 10 is omitted. If more than one monochromatic group is omitted, the solution is not considered correct overall.

-3 points. Not considered that both colors must be present, and not considered the alternative coloring.
Task 2. Find the number of ways to color all natural numbers from 1 to 20 in blue and red such that both colors are used and the product of all red numbers is coprime with the product of all blue numbers.