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# Let the function p(n) represent the product of the first n p

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Intern
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Let the function p(n) represent the product of the first n p [#permalink]

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28 Jul 2014, 02:57
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Question Stats:

58% (01:33) correct 42% (01:37) wrong based on 196 sessions

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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
[Reveal] Spoiler: OA

Kudos [?]: 74 [0], given: 60

Math Expert
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Let the function p(n) represent the product of the first n p [#permalink]

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28 Jul 2014, 03:43
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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.

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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14 Sep 2014, 13:41
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.

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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02 May 2015, 02:53
I did not get how to choose between ii and iii. We know x to be prime. How can we dismiss ii then?

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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01 Jun 2015, 14:03
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

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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27 Oct 2016, 09:10
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

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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27 Oct 2016, 09:35
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.
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Re: Let the function p(n) represent the product of the first n p [#permalink]

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27 Oct 2016, 09:45
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?

Kudos [?]: 11 [0], given: 123

Math Expert
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Posts: 43334

Kudos [?]: 139638 [0], given: 12794

Let the function p(n) represent the product of the first n p [#permalink]

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27 Oct 2016, 09:50
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.
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Re: Let the function p(n) represent the product of the first n p [#permalink]

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27 Oct 2016, 09:59
Got it, thanks Bunuel.

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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27 Oct 2016, 15:45
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

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Re: Let the function p(n) represent the product of the first n p [#permalink]

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23 Dec 2017, 18:55
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