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For any subset $S \subseteq {1, 2, 3, ...., 15},$ a number n is called a “twister” for S if n and $n + | S |$ are both members of S , where $| S |$ denotes the number of members of S . The average number of twisters over all possible subsets $S \subseteq {1, 2, 3, ...., 15}$ is $\frac{a}{b} (G C D (a, b) = 1)$ , then which of the following is/are TRUE?

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$| b - a |$ is a multiple of 5

a + b is even

a + b is odd

a + b is prime number

answer is A, C.

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Detailed Solution

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Explanation:

A "twister" is a number $ n $ such that $ n \in S $ and $ n + |S| \in S $.
$ S \subseteq \{1, 2, 3, \dots, 15\} $, and the total subsets are $ 2^{15} $.

The average number of twisters is calculated as $ \frac{a}{b} $, where $ \text{GCD}(a, b) = 1 $.

Analysis of Options:

Option a: $ |b - a| $ is a multiple of 5 → True.
Option b: $ a + b $ is even → False.
Option c: $ a + b $ is odd → True.
Option d: $ a + b $ is prime → False.

Final Answer: The correct options are a and c.

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