Question:

Step 1: Identify the set $S = \{14, 15, 16, 17, 18, 19, 20, 21, 22\}$, which has 9 elements. Step 2: For a subset with two or more elements, the least element $a$ and greatest element $b$ satisfy $a < b$. The sum $a + b$ must have an odd number of divisors, which means $a + b$ must be a perfect square. Step 3: List all possible pairs $(a, b)$ with $a, b \in S$, $a < b$, and $a + b$ is a perfect square. The possible sums range from $14+15=29$ to $21+22=43$. Perfect squares in this range: $36$ (since $6^2=36$). Check: $a + b = 36$: pairs are $(14,22), (15,21), (16,20), (17,19)$. Also $a+b = 25$: not possible since min sum is 29. $a+b=49$: not possible since max is 43. So only $a+b=36$. Step 4: For each valid pair $(a,b)$, count subsets where $a$ is the minimum and $b$ is the maximum. The elements strictly between $a$ and $b$ in $S$ can be any subset of the elements of $S$ strictly between $a$ and $b$. - Pair $(14,22)$: elements between them in $S$: $\{15,16,17,18,19,20,21\}$ — 7 elements, giving $2^7 = 128$ subsets. - Pair $(15,21)$: elements between them in $S$: $\{16,17,18,19,20\}$ — 5 elements, giving $2^5 = 32$ subsets. - Pair $(16,20)$: elements between them in $S$: $\{17,18,19\}$ — 3 elements, giving $2^3 = 8$ subsets. - Pair $(17,19)$: elements between them in $S$: $\{18\}$ — 1 element, giving $2^1 = 2$ subsets. Step 5: Total $N = 128 + 32 + 8 + 2 = 170$. Step 6: Factorize $N = 170 = 2 \times 5 \times 17$. Number of prime factors = 3 (option a is correct). Number of divisors $= (1+1)(1+1)(1+1) = 8$ (option b is correct). Sum of digits of $N = 1 + 7 + 0 = 8$ (option c is correct). Number of divisors is 8, not 16 (option d is incorrect).