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30 Jul 2017, 22:36
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (02:24) correct
39%
(02:08)
wrong
based on 191
sessions
History
Date
Time
Result
Not Attempted Yet
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.
(1) p is the least possible integer which satisfies (3^p + 2) is NOT a prime number.
(2) 3^p + 2 < 500.
31 Jul 2017, 05:06
Kudos
Bookmarks
Bunuel
1) If p= 1.........(3^p + 2) =5 = Prime number...invalid
p= 2.........(3^p + 2) =11 = Prime number...invalid
p= 3.........(3^p + 2) =29 = Prime number...invalid
p= 4.........(3^p + 2) =83 = Prime number...invalid
p= 5.........(3^p + 2) =245 = NOT Prime number...valid
Then the reminder of 245/5 is zero
Sufficient
2) 3^p + 2 < 500
Use the numbers from above p =5 and reminder is zero.
Note that p can NOT equal to 6, otherwise it will be greater than 500.
Sufficient
Answer: D
01 Aug 2017, 01:59
Kudos
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Bunuel
Given p>0.
p=1 ==> \((3^1 + 2)\) = 5 --> prime
p=2 ==> \((3^2 + 2)\) = 11 --> prime
p=3 ==> \((3^3 + 2)\) = 30 --> not prime
p=4 ==> \((3^4 + 2)\) = 83 --> prime
p=5 ==> \((3^5 + 2)\) = 245 --> not prime
p=6 ==> \((3^6 + 2)\) = 731 --> prime
Quote:
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:
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
