Bunuel wrote:

p is the value of the positive integer such that (3^p + 2) is NOT a prime number. What is the remainder when (3^p + 2) is divided by p?

(1) p is the least possible integer which satisfies (3^p + 2) is NOT a prime number.

(2) 3^p + 2 < 500.

Given p>0.

p=1 ==> \((3^1 + 2)\) = 5 --> prime

p=2 ==> \((3^2 + 2)\) = 11 --> prime

p=3 ==> \((3^3 + 2)\) = 30 --> not primep=4 ==> \((3^4 + 2)\) = 83 --> prime

p=5 ==> \((3^5 + 2)\) = 245 --> not primep=6 ==> \((3^6 + 2)\) = 731 --> prime

**Quote:**

(1) p is the least possible integer which satisfies (3^p + 2) is NOT a prime number.

From the above list p=3 is the least possible integer which satisfies \((3^p + 2)\) is NOT a prime.

\((3^p + 2)\)/p

p=3 ==> \((3^3 + 2)\) = 30

30/3 ==> remainder=0

Sufficient

**Quote:**

(2) 3^p + 2 < 500.

From the list above, p=3 and p=5 satisfies statement 2.

\((3^3 + 2)\) = 30 --> 30/3 ==> remainder= 0

\((3^5 + 2)\) = 245 --> 245/5 ==> remainder= 0

Value of p can't go beyond 5 as for p=6, we get \((3^6 + 2)\) = 731 >500

Thus, for both p=3 and p=5 the remainder is 0.

Sufficient

Answer: D

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Thanks,

Ramya