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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # p is the value of the positive integer such that (3^p + 2) is NOT a

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Math Expert V
Joined: 02 Sep 2009
Posts: 59729
p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 54% (02:23) correct 46% (02:11) wrong based on 90 sessions

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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.

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SVP  V
Joined: 26 Mar 2013
Posts: 2345
Re: p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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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

Intern  B
Joined: 09 Dec 2014
Posts: 37
Re: p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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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

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Ramya
Intern  B
Joined: 15 May 2015
Posts: 13
GPA: 3.5
Re: p is the value of the positive integer such that (3^p + 2) is NOT a  [#permalink]

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Ramyanag wrote:
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

29 not 30 Re: p is the value of the positive integer such that (3^p + 2) is NOT a   [#permalink] 08 Dec 2019, 12:18
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