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p is the value of the positive integer such that (3^p + 2) is NOT a

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p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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New post 30 Jul 2017, 22:36
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D
E

Difficulty:

  65% (hard)

Question Stats:

55% (02:29) correct 45% (02:10) wrong based on 72 sessions

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Re: p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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New post 31 Jul 2017, 05:06
2
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.



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
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Re: p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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New post 01 Aug 2017, 01:59
1
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 prime
p=4 ==> \((3^4 + 2)\) = 83 --> prime
p=5 ==> \((3^5 + 2)\) = 245 --> not prime
p=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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Ramya
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Re: p is the value of the positive integer such that (3^p + 2) is NOT a   [#permalink] 01 Aug 2017, 01:59
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