Question

How many ways are there to color the numbers 1, 2, 3, 4, 5, 6 with the colors red, green and blue such that no number is colored the same as one of its proper divisors? (The proper divisors of a number are the divisors that are not equal to the number itself)​

Answer 1

Given: Numbers to color = 1, 2, 3, 4, 5, 6

Colors = red, green and blue  (i.e. 3 colors)

To find : The number of ways  to color the numbers 1, 2, 3, 4, 5, 6 with the colors red, green and blue such that no number is colored the same as one of its proper divisors.

Case 1) Taking prime number can take any of three colors, except one of 2 and 3.

Then 2 and 5 can have $3\times 3 = 9$ways

Case 2) Taking Multiples of 2.

4 can take any of the remaining 2 colors, while 6 would take any other cor.

Thus number of ways $2\times1=2$

Case 3) Taking Multiples of 3.

6 is a multiple of 3.

Number of ways =$2\times 1 = 2$

Total ways = $9 \times 2 \times 2 = 36$

Hence, there are 36 ways to color the numbers 1, 2, 3, 4, 5, 6 with the colors red, green and blue such that no number is colored the same as one of its proper divisors.

Answer 2

Answer:

answer is 36 please follow me