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If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 10:30
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 12:06
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A is ans on solving a we get 1+ n^3 = P^2 so it will be satisfied with n=2 and p=3.. the reason is..p will be alws greater than n..so p will alws be odd no..so its square will be alws odd... so n^3 must be even which is possible with only prime no 2.. so 1+n^3 will bcome odd... so satisfied...
2) on solving eq bcom 8p^2 +1 =m..so I m unable to find any prime no which is 24k+1..where k is any + integer.. so not satisfied



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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 12:11
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Bunuel wrote: If \(p\) is a prime number, what is the value of \(p\)?
(1) \(\sqrt[3]{1p^2}=n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number. (1) \(\sqrt[3]{1p^2}=n\) OR \(p^2=1+n^3=(1+n)(n^2n+1)\) If n=2, p=3 If n>2, then n is odd, (n+1) is even, hence p^2 is even, but this is only possible if p=2 and if p=2, n is cube_root(3) which is not an integer So only possibility is p=3 Sufficient (2) \(8p^2+1=m\) p is an integer and a prime so it can either be of the form 3k+1 or 3k1 or be 3 (all other numbers of form 3k are composite) 8*(3k+1)^2+1=8*(9k^2+6k+1)+1=3*(24k^2+16k+3) .. which cannot be a prime since k>=1 8*(3k1)^2+1=8*(9k^26k+1)+1=3*(24k^216k+3) .. which cannot be a prime since k>=1 8*3^2+1=73 is a prime So only possible value of p is 3 Sufficient Answer is (D)
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 12:27
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hey Man p is an integer and a prime so it can either be of the form 6k+1 or 6k1... not 3k+1 and 3k1



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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 12:29
sudhanshushankerjha wrote: hey Man p is an integer and a prime so it can either be of the form 6k+1 or 6k1... not 3k+1 and 3k1 You can always divide all integers into 3 sets : 3k , 3k+1 , 3k1 In case of primes, it is easy to see only one element in the first set, i.e., 3
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 12:43
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ya agreed D is ans..Thnx



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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 15:32
Bunuel wrote: If \(p\) is a prime number, what is the value of \(p\)?
(1) \(\sqrt[3]{1p^2}=n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number. I wrote this question, so below is my solution: (1) \(\sqrt[3]{1p^2}=n\), where \(n\) is a prime number. It's clear that \(p\neq{2}\) as in this case \(\sqrt[3]{1p^2}\neq{integer}\), so \(p\) is a prime more than 2, so odd. Then \(1p^2=oddodd=even\) > cube root from even is either an even integer or not an integer at all, we are told that \(\sqrt[3]{1p^2}=n=integer\), so \(n=even=prime\) > \(n=2\) (the only even prime) > \(\sqrt[3]{1p^2}=2\) > \(p=3\). Sufficient. (2) \(8p^2+1=m\), where \(m\) is a prime number. \(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) > \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) > so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\). Sufficient. Answer: D.
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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03 Oct 2010, 22:08
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Statement 2: take prime = 6k+1 and 6k1 \(8*((6k+1)^2 + 1) = 8*36k^2 + 12k +9 =\) 3* something => not a prime \(8*((6k1)^2 + 1) = 8*36k^2  12k +9 = 3*\) something => not a prime so any prime number of the form 6k+1 and 6k1 is not possible. lowest prime of the above form is for k=1 i.e. 5. Thus only 2 possibilities p=2 and p=3. Only p=3 satisfies. BUNUEL Great question.
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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05 Oct 2010, 00:01
great question Bunuel



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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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05 Oct 2010, 10:55
I am really confused. Can someone please explain me, Statement 2: I understand that if we work out the 8P^2 + 1 : where P = 2 we get m = 33, which is NOT a prime. Great! I am onboard till here. Is the strategy to try every prime number from 2, onwards?? How is the value of P chosen?? You state P cannot be any prime but 3 for 8P^2 + 1 to be a prime number. How can you be so sure? I do not have an exhaustive list of all prime number, but this claim is little far fetched. Please clear my confusion.
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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05 Oct 2010, 11:04
vrajesh wrote: I am really confused. Can someone please explain me,
Statement 2:
I understand that if we work out the 8P^2 + 1 : where P = 2 we get m = 33, which is NOT a prime.
Great! I am onboard till here.
Is the strategy to try every prime number from 2, onwards??
How is the value of P chosen??
You state P cannot be any prime but 3 for 8P^2 + 1 to be a prime number. How can you be so sure? I do not have an exhaustive list of all prime number, but this claim is little far fetched.
Please clear my confusion. There are infinitely many prime numbers so there is no "list of all primes". Next, \(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 ( not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) > \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) > so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\).
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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05 Oct 2010, 12:47
Hi Bunuel, OK. But I am still not clear about why is it important to be divisible by 3. "because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 ( not a multiple of 3 when squared yields remainder of 1 when divided by 3)" Bunuel wrote: There are infinitely many prime numbers so there is no "list of all primes".
Next, \(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) > \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) > so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\). I guess, you will have to explain this a little more.
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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16 Oct 2010, 07:58
Raths wrote: tough one... Yes nice question 700+
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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23 Oct 2010, 04:28
D ...intially i thought it as A coz i was wrongly marked 57 as prime Silly mistake :0



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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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23 Oct 2010, 19:32
Tough one, I could solve the question by plugging in the numbers but didn't think of trying (especially for 2nd statement) the above concepts, despite knowing for a number to be prime it should be either \(6k+1\) or \(6k1\). Similarly, for statemet1 despite knowing that \(p\) can not be \(2\) and it would be more than didn't think that it must be odd and apply Odd & Even concept here. Thank you! for reminding me to use concepts rather than just guessing and then getting doubtful solutions (and doing silly mistakes).
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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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20 Jul 2016, 11:18
Bunuel wrote: If \(p\) is a prime number, what is the value of \(p\)?
(1) \(\sqrt[3]{1p^2}=n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number. Please find the solution as attached. Answer: option D
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If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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21 Jul 2016, 09:16
GMATinsight wrote: Bunuel wrote: If \(p\) is a prime number, what is the value of \(p\)?
(1) \(\sqrt[3]{1p^2}=n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number. Please find the solution as attached. Answer: option D Dear GMATinsight, Thanks for you explanation. I did the same as you did in solving those question. But I had a doubt that there maybe a higher prime number that satisfy fact 2 so I kept doing this to 13 & 17. How should I be confident to stop at 11 like you?



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Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]
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21 Jul 2016, 20:14
Mo2men wrote: GMATinsight wrote: Bunuel wrote: If \(p\) is a prime number, what is the value of \(p\)?
(1) \(\sqrt[3]{1p^2}=n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number. Please find the solution as attached. Answer: option D Dear GMATinsight, Thanks for you explanation. I did the same as you did in solving those question. But I had a doubt that there maybe a higher prime number that satisfy fact 2 so I kept doing this to 13 & 17. How should I be confident to stop at 11 like you? Mo2menTo be Confident 1) You need to find out a pattern in solutions because there are always patterns followed among all acceptable or unacceptable solutions 2) To find a pattern you need three instances/examples to observe I saw the pattern in solutions that each of the solutions that I was finding was a multiple of 3 so I concluded that this pattern is going to continue Trusting the pattern based on three instances usually gets you the correct answer in 99% cases and the remaining 1% are usually made by Manhattan intentionally to prove this thought process incorrect. Thankfully Manhattan doesn't make GMAT so the chances are bleak that you get wrong answer trusting the pattern
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