Question

A is a set of first 10 natural numbers. B is a subset of A such that B consists of exactly 2 numbers, both co-primes to each other. Find the maximum number of such possible subsets.

• A. 29
• B. 28
• C. 33
• D. 31

Answer 1

The correct option is D 31

For 10, number of co-primes = 10*(1/2)*(4/5) = 4
For 9, number of co-primes = 9*(2/3) = 6
For 8, number of co-primes = 8*(1/2) = 4
For 7, number of co-primes = 7*(6/7) = 6
For 6, number of co-primes = 6*(1/2)*(2/3) = 2
For 5, number of co-primes = 5*(4/5) = 4
For 4, number of co-primes = 4*(1/2) = 2
For 3, number of co-primes = 3*(2/3) = 2
For 2, number of co-primes = 2*(1/2) = 1
Total number of co-primes = 4 + 6 + 4 + 6 + 2 + 4 + 2 + 2 + 1 = 31

Alternatively,
In subset B, if one number is 1, then, others would be 2, 3, 4...10, which are 9 numbers.
When one number is 2, then other numbers would be 3, 5, 7, 9 (which is 4 numbers).
When one number is 3, then other numbers would be 4, 5, 7, 8. 10 (which is 5 numbers).
When one number is 4, then other numbers are 5, 7, 9 (which are three numbers).
When one number is 5, then other numbers would be 6, 7, 8. 9 (which is 4 numbers).
When one number is 6, then other number would be 7 (which is one number).
When one number is 7, then other numbers would be 8, 9, 10 (which are three numbers.
When one number is 8, then other number would be 9 (which is one number).
When one number is 9, then other number would be 10 (which is one number).
Therefore, total number of subsets = 9 + 4 + 5 + 3 + 4 + 1 + 3 + 1 + 1 = 31