+274 How many subsets of the set of divisors of 72 contain only composite numbers? For example, {8, 9} and {4, 8, 12} are two such sets. Include the empty set in your count.
To determine the number of subsets of the set of divisors of 72 that contain only composite numbers, we first need to find the divisors of 72.
The prime factorization of 72 is 2^3 * 3^2, which means its divisors are all the possible combinations of the prime factors raised to various powers. The divisors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
Now, let's count the number of subsets that contain only composite numbers. Since the empty set is also considered a subset, we need to include it in our count.
Out of the 12 divisors, 1 is not a composite number, so we cannot include it in any subset that consists only of composite numbers. This means we have 11 remaining divisors to choose from.
For each divisor, we have two choices: include it in a subset or exclude it. Therefore, the total number of subsets that contain only composite numbers is 2^11.
Including the empty set, the total count is 2^11 + 1 = 2049.
So, there are 2049 subsets of the set of divisors of 72 that contain only composite numbers, including the empty set.
There are 12 divisors of 72 as follows:
(1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72)
There are: 2^12 ==4,096 subsets including the empty set.
There are: 2^11 + 2^10 ==3,072 subsets that have either a"2" or a "3" or both in them.
Therefore: There are: 4,096 - 3,072 ==1,024 subsets that contain only composite numbers.
+274 Their two different solutions here and I'm not sure which one seems correct. Can someone else pls break the tie?
