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Intern  Joined: 14 Jan 2014
Posts: 11
Let the function p(n) represent the product of the first n p  [#permalink]

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4 00:00

Difficulty:   65% (hard)

Question Stats: 57% (02:14) correct 43% (02:13) wrong based on 222 sessions

### HideShow timer Statistics Let the function p(n) represent the product of the first n prime numbers, where n > 0. If x = p(n) + 1, which of the following must be true?

(i) x is always odd

(ii) x is always prime

(iii) x is never the square of an integer

A. ii only
B. iii only
C. i and ii only
D. i and iii only
E. ii and iii only
Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Let the function p(n) represent the product of the first n p  [#permalink]

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2
1
Vijayeta wrote:
Let the function p(n) represent the product of the first n prime numbers, where n > 0. If x = p(n) + 1, which of the following must be true?

(i) x is always odd

(ii) x is always prime

(iii) x is never the square of an integer

A. ii only
B. iii only
C. i and ii only
D. i and iii only
E. ii and iii only

p(n) is always even, because the first prime is 2 and no matter what n is, 2 always will be a divisor of p(n). Thus, p(n) + 1 = even + 1 = odd. So, (i) is always true.

Now, use logic:

If (ii) is true (so if x is always prime), then (iii) must automatically be true: no prime is the square of an integer. So, the correct answer must be i only; i, ii, and iii only; or i and iii only. since only "i and iii only" is among the options, then it must be true.

Or, since (i) is always true, then from options the answer must be either C or D. C cannot be correct because if (ii) is true, then so must be (iii). Thus only D remains.

_________________
Manager  Joined: 29 May 2013
Posts: 55
GMAT 1: 710 Q49 V38 GPA: 4
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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Bunuel wrote:
Vijayeta wrote:
Let the function p(n) represent the product of the first n prime numbers, where n > 0. If x = p(n) + 1, which of the following must be true?

(i) x is always odd

(ii) x is always prime

(iii) x is never the square of an integer

A. ii only
B. iii only
C. i and ii only
D. i and iii only
E. ii and iii only

p(n) is always even, because the first prime is 2 and no matter what n is, 2 always will be a divisor of p(n). Thus, p(n) + 1 = even + 1 = odd. So, (i) is always true.

Now, use logic:

If (ii) is true (so if x is always prime), then (iii) must automatically be true: no prime is the square of an integer. So, the correct answer must be i only; i, ii, and iii only; or i and iii only. since only "i and iii only" is among the options, then it must be true.

Or, since (i) is always true, then from options the answer must be either C or D. C cannot be correct because if (ii) is true, then so must be (iii). Thus only D remains.

A nice approach.
I used the same approach. It took me 1 min 51 sec. I wonder if I could have figured the approach sooner though.
Intern  Joined: 23 Nov 2014
Posts: 31
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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I did not get how to choose between ii and iii. We know x to be prime. How can we dismiss ii then?
Manager  Joined: 07 Dec 2009
Posts: 86
GMAT Date: 12-03-2014
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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Gmatdecoder wrote:
I did not get how to choose between ii and iii. We know x to be prime. How can we dismiss ii then?

You use the answer choices to make your decision. If ii is true than iii has to be true but we don't have any answer choice with all 3 options. Hence we go for i & iii
Intern  B
Joined: 16 Mar 2014
Posts: 16
GMAT Date: 08-18-2015
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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Bunuel wrote:
Vijayeta wrote:
Let the function p(n) represent the product of the first n prime numbers, where n > 0. If x = p(n) + 1, which of the following must be true?

(i) x is always odd

(ii) x is always prime

(iii) x is never the square of an integer

A. ii only
B. iii only
C. i and ii only
D. i and iii only
E. ii and iii only

p(n) is always even, because the first prime is 2 and no matter what n is, 2 always will be a divisor of p(n). Thus, p(n) + 1 = even + 1 = odd. So, (i) is always true.

Now, use logic:

If (ii) is true (so if x is always prime), then (iii) must automatically be true: no prime is the square of an integer. So, the correct answer must be i only; i, ii, and iii only; or i and iii only. since only "i and iii only" is among the options, then it must be true.

Or, since (i) is always true, then from options the answer must be either C or D. C cannot be correct because if (ii) is true, then so must be (iii). Thus only D remains.

Hi Bunuel,
I'm just wondering whether ii is not true? Is there any case that makes ii is not true?

Thanks indeed
Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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yenh wrote:
Bunuel wrote:
Vijayeta wrote:
Let the function p(n) represent the product of the first n prime numbers, where n > 0. If x = p(n) + 1, which of the following must be true?

(i) x is always odd

(ii) x is always prime

(iii) x is never the square of an integer

A. ii only
B. iii only
C. i and ii only
D. i and iii only
E. ii and iii only

p(n) is always even, because the first prime is 2 and no matter what n is, 2 always will be a divisor of p(n). Thus, p(n) + 1 = even + 1 = odd. So, (i) is always true.

Now, use logic:

If (ii) is true (so if x is always prime), then (iii) must automatically be true: no prime is the square of an integer. So, the correct answer must be i only; i, ii, and iii only; or i and iii only. since only "i and iii only" is among the options, then it must be true.

Or, since (i) is always true, then from options the answer must be either C or D. C cannot be correct because if (ii) is true, then so must be (iii). Thus only D remains.

Hi Bunuel,
I'm just wondering whether ii is not true? Is there any case that makes ii is not true?

Thanks indeed

$$p(6) + 1 = 2*3*5*7*11*13+1=30031 = 59*509$$ is not a prime.
_________________
Intern  B
Joined: 16 Mar 2014
Posts: 16
GMAT Date: 08-18-2015
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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Quote:

Hi Bunuel,
I'm just wondering whether ii is not true? Is there any case that makes ii is not true?

Thanks indeed

Quote:
$$p(6) + 1 = 2*3*5*7*11*13+1=30031 = 59*509$$ is not a prime.

Thanks for your quick response. Btw, is there any way to be sure (theoretically) in case we cannot figure out by example?
Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Let the function p(n) represent the product of the first n p  [#permalink]

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yenh wrote:
Quote:

Hi Bunuel,
I'm just wondering whether ii is not true? Is there any case that makes ii is not true?

Thanks indeed

Quote:
$$p(6) + 1 = 2*3*5*7*11*13+1=30031 = 59*509$$ is not a prime.

Thanks for your quick response. Btw, is there any way to be sure (theoretically) in case we cannot figure out by example?

There is no known formula for prime numbers (in fact it's one of the biggest math challenges), so p(n) + 1 cannot be prime for all values of n, else we would have the formula which gives primes.
_________________
Intern  B
Joined: 16 Mar 2014
Posts: 16
GMAT Date: 08-18-2015
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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Got it, thanks Bunuel.
Current Student B
Joined: 26 Jan 2016
Posts: 100
Location: United States
GPA: 3.37
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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try plugging in numbers. If n=2 then 2X3+1=7
if n=3 then 2*3*5=30+1=31 the answer will always be odd b/c 2 is a prime and the product will always be an even plus 1

If you try n=4 then you get 2*3*5*7=210 which is not a prime

D
Non-Human User Joined: 09 Sep 2013
Posts: 11666
Re: Let the function p(n) represent the product of the first n p  [#permalink]

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_________________ Re: Let the function p(n) represent the product of the first n p   [#permalink] 23 Dec 2017, 19:55
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