First, $$2^6 = 64$$, so we need primes between 64 and 84, inclusive. Let's check each number systematically: 67 (prime), 71 (prime), 73 (prime), 79 (prime), 83 (prime). All other numbers in this range are composite. For example: 64 = $$2^6$$, 65 = 5×13, 69 = 3×23, 77 = 7×11, 81 = $$3^4$$. Therefore, there are exactly 5 prime numbers in this range.

Let the two primes be $$p$$ and $$q$$ where $$p + q = 30$$ and one prime is greater than 20. The primes greater than 20 and less than 30 are: 23, 29. If $$p = 23$$, then $$q = 30 - 23 = 7$$, and 7 is prime. If $$p = 29$$, then $$q = 30 - 29 = 1$$, but 1 is not considered prime. Therefore, there is only one pair: (23, 7). Choice B would suggest there are 2 pairs, but the second possibility doesn't work. Choices C and D overcount non-existent pairs.

We need to find the smallest positive integer $$n$$ such that $$2n + 1$$ is composite. Let's check values: For $$n = 1$$: $$2(1) + 1 = 3$$ (prime). For $$n = 2$$: $$2(2) + 1 = 5$$ (prime). For $$n = 3$$: $$2(3) + 1 = 7$$ (prime). For $$n = 4$$: $$2(4) + 1 = 9 = 3^2$$ (composite). So the smallest value of $$n$$ is 4, making $$2n + 1 = 9$$. Choice B (15) corresponds to $$n = 7$$, choice C (21) to $$n = 10$$, and choice D (25) to $$n = 12$$, all of which are larger than $$n = 4$$.

We're looking for twin prime pairs where both $$k$$ and $$k + 2$$ are prime, with $$3 < k < 50$$. The twin prime pairs in this range are: (5,7), (11,13), (17,19), (29,31), and (41,43). Therefore, the possible values of $$k$$ are: 5, 11, 17, 29, and 41. This gives us exactly 5 possible values.

First, let's calculate $$m$$ and $$n$$: $$m = 2^3 \cdot 3^2 \cdot 5 = 8 \cdot 9 \cdot 5 = 360$$ and $$n = 2^2 \cdot 3 \cdot 7 = 4 \cdot 3 \cdot 7 = 84$$. Therefore, $$m + n = 360 + 84 = 444$$. Since $$444 = 4 \cdot 111 = 4 \cdot 3 \cdot 37 = 12 \cdot 37$$, we see that 444 is composite. Choice A is wrong because we have specific values, not a general statement. Choice B is incorrect because the sum of composite numbers isn't necessarily composite. Choice D is wrong because we can determine the exact nature of this specific sum.

To determine if 91 is prime or composite, we need to check for factors. $$91 = 7 \times 13$$, so 91 is composite. Choice A is incorrect because checking divisibility by 2, 3, 5, 7 is insufficient - we need to check all primes up to $$\sqrt{91} \approx 9.5$$, and 91 is divisible by 7. Choice C is wrong because 91 IS divisible by 7 (and 7 is between 2 and 9). Choice D is incorrect because 91 is not divisible by 3 (sum of digits: 9 + 1 = 10, which is not divisible by 3). The correct answer is B because $$7 \times 13 = 91$$.

If $$N = 2^a \cdot 3^b$$, then the number of divisors is $$(a+1)(b+1) = 6$$. The factor pairs of 6 are (1,6), (2,3), (3,2), and (6,1), giving us $$(a,b) = (0,5), (1,2), (2,1), (5,0)$$. The corresponding values are: $$N = 3^5 = 243$$, $$N = 2^1 \cdot 3^2 = 18$$, $$N = 2^2 \cdot 3^1 = 12$$, and $$N = 2^5 = 32$$. Since $$N > 12$$, we exclude $$N = 12$$. Among the remaining candidates (18, 32, 243), the smallest is 18.

We need to find when $$n^2 + n + 41$$ is composite. Let's test values systematically. When $$n = 40$$: $$40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41^2$$, which is composite. We can verify that for smaller values like $$n = 39$$: $$39^2 + 39 + 41 = 1521 + 39 + 41 = 1601$$, which is prime. Choice B (41) gives $$41^2 + 41 + 41 = 41(41 + 1 + 1) = 41 imes 43$$, which is also composite, but 41 > 40. Choice C (80) gives a composite number but is larger than 40. Choice D (81) also works but is larger.