Let A={1,2,3,4,5,6,7}. Define B={T ⊆ A : either 1 ∉ T or 2 ∈ T} and C={T ⊆ A: T the sum of all the elements of T is a prime number }. Then the number of elements in the set B ∪ C is ________________.

Solution

Maths sir · 3 steps

B is almost everything, so count what B leaves out and add back only those with a prime sum.

1. Size of B. A subset fails to be in B only when 1 ∈ T and 2 ∉ T. The other 5 elements are free, so 2⁵ = 32 subsets lie outside B, and |B| = 2⁷ - 32 = 96.

   |B ∪ C| = |B| + | {T ∉ B : sum of T is prime} |

2. Subsets outside B. Each such subset is T = {1} ∪ S with S ⊆ {3,4,5,6,7}, so its sum is 1 + (sum of S). The sum of S runs from 0 to 25, so the sum of T is at most 26. We need 1 + (sum of S) to be prime.

   Sum of T (prime)	Sum of S	Possible S	Count
   2 or 3	1 or 2	none	0
   5	4	{4}	1
   7	6	{6}	1
   11	10	{3,7}, {4,6}	2
   13	12	{5,7}, {3,4,5}	2
   17	16	{4,5,7}, {3,6,7}	2
   19	18	{5,6,7}, {3,4,5,6}	2
   23	22	{4,5,6,7}	1

3. So. There are 1+1+2+2+2+2+1 = 11 subsets outside B with a prime sum. So |B ∪ C| = 96 + 11 = 107.

The common slip: 96. Assuming every prime-sum subset already lies in B and stopping at |B| gives 96. The 11 prime-sum subsets that contain 1 but not 2 still have to be added.

The idea to keep. When a set is defined by an "either ... or" condition, count its complement. It is usually small, and only that small leftover needs to be checked against the second set.

Answer from the NTA final answer key. Solution by PrepWiser.