The function (3 Δ 47) equals the product (3)(5)(7)…(43)(45)(47). This product is a very large odd number, as it is the product of only odd numbers and thus does not have 2 as a factor. Therefore, (3 Δ 47) + 2 = Odd + Even = Odd, and (3 Δ 47) + 2 does not have 2 as a factor either.
Every odd prime number between 3 and 47 inclusive is a factor of (3 Δ 47), since each of these primes is a component of the product. For example, (3 Δ 47) is divisible by 3, since dividing by 3 yields an integer — the product (5)(7)(9)…(43)(45)(47). Now, consider the sum (3 Δ 47) + k, where k is an integer. The sum will only be divisible by 3 if k is
also divisible by 3. In other words, when we divide (3 Δ 47) + k by 3, we are evaluating (3 Δ 47)/3 + k/3. Because (3 Δ 47)/3 is an integer, k/3 must also be an integer to yield an integer sum.
In this problem, k = 2, which is not divisible by any of the odd primes between 3 and 47. Since (3 Δ 47) IS divisible, but 2 is NOT divisible, we conclude that the sum (3 Δ 47) + 2 is NOT divisible by any of the odd primes between 3 and 47. So, (3 Δ 47) + 2 is not divisible by any prime number less than or equal to 47. The smallest prime
factor of (3 Δ 47) + 2 must be greater than 47. Thus, the minimum possible prime factor is 53, since that is the smallest prime greater than 47.
The correct answer is A.