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If N is a positive odd integer, is N prime?
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27 Jul 2017, 08:02
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If N is a positive odd integer, is N prime?
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27 Jul 2017, 09:00
carcass wrote: If N is a positive odd integer, is N prime?
(1) \(N = 2^k+ 1\) for some positive integer k.
(2) N + 2 and N + 4 are both prime. hi... lets see the statements.. (1) \(N = 2^k+ 1\) for some positive integer k. if k = 2, N = \(2^2+1=5\).. YES if k= 3, N=\(2^3+1=9\)... No Insuff (2) N + 2 and N + 4 are both primeif N+2 and N+4 are prime, ONE of N or N+2 or N+4 will surely be MULTIPLE of 3..so N can be prime only when N=3, otherwise always NOInsuff combined Nothing new E
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Re: If N is a positive odd integer, is N prime?
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27 Jul 2017, 09:08
carcass wrote: If N is a positive odd integer, is N prime?
(1) \(N = 2^k+ 1\) for some positive integer k.
(2) N + 2 and N + 4 are both prime. Is the answer C? from 1, k=1,2 satisfy, but k=3 doesn't. insuff from 2, N=1, 3 satisfy, but 1 is neither prime nor composite. insuff together, N =3 satisfies both the criteria.
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If N is a positive odd integer, is N prime?
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27 Jul 2017, 09:12
rekhabishop wrote: carcass wrote: If N is a positive odd integer, is N prime?
(1) \(N = 2^k+ 1\) for some positive integer k.
(2) N + 2 and N + 4 are both prime. Is the answer C? from 1, k=1,2 satisfy, but k=3 doesn't. insuff from 2, N=1, 3 satisfy, but 1 is neither prime nor composite. insuff together, N =3 satisfies both the criteria. Answer will be E , N can be 3 or 9.



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If N is a positive odd integer, is N prime?
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08 Jan 2018, 07:18
If N is a positive odd integer, is N prime?
(1) N=2^k+1N for some positive integer k.N can be 3, 5, or 9 using values of k = 1, 2 and 3 respectively INSUFF(2) N + 2 and N + 4 are both prime.If N = 3 then N + 2 = 5 and N + 4 = 7 COMPLIES (we pick the first prime of the examples used for S1). If N = 9 then N + 2 = 11 and N + 4 = 13 COMPLIES (we pick the first nonprime of the examples used for S1). INSUFF(1)(2) Same examples fulfill both Ss (in this case the prime 3 and the nonprime 9). INSUFFAC: E
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Re: If N is a positive odd integer, is N prime?
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15 Jun 2018, 22:45
chetan2u wrote: carcass wrote: If N is a positive odd integer, is N prime?
(1) \(N = 2^k+ 1\) for some positive integer k.
(2) N + 2 and N + 4 are both prime. hi... lets see the statements.. (1) \(N = 2^k+ 1\) for some positive integer k. if k = 2, N = \(2^2+1=5\).. YES if k= 3, N=\(2^3+1=9\)... No Insuff (2) N + 2 and N + 4 are both primeif N+2 and N+4 are prime, ONE of N or N+2 or N+4 will surely be MULTIPLE of 3..so N can be prime only when N=3, otherwise always NOInsuff combined Nothing new E Responding to a PM ... Why should one of n, n+2 or n+4 be a multiple of 3.... If n is odd, all three will be odd.... 1,3,5 or 3,5,7.... In these 3 is multiple of 3 Next three are 5,7,9..., So here 3 has moved out of set but 9 has come in Say n is even 2,4,6 or 4,6,8 or 6,8,10.... Here 6 is present Next would be 8,10,12.... So 6 has moved out but 12 has come in... The reason for this is the multiple of 3 comes after 3.. Similarly if you are looking for say n,n+2,.....n+12 these are 7 terms and any one of them would surely be multiple of all odd prime numbers till 7... 3,5,7
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Re: If N is a positive odd integer, is N prime?
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11 Sep 2018, 13:51
Hi, First thankyou chetan2u, you have given a wonderful ready to use result that would be very useful for lot of questions especially on divisibility. If i may take liberty to repost the result that you have shared, if I were to remember ( actuality no need to memories since is understood the reasoning behind it ) this result it would be as follows If we have a consecutive series of "n" odd or "n"even numbers , then one of them will be will be definitely divisible by odd numbers <=n. Lets take an series of 5 consecutive odd numbers, then as per this one of them will be definitely divisible by odd numbers <=5 ( that means the one of the numbers will be definitely divisible by (3, 5) say the series is 5 consecutive odd numbers 101, 103,105, 107,109 So we have 105 is divisible by 5& 3 and 109 is divisible by 3. Lets take another series of 5 consecutive even numbers, then as per this one of them will be definitely divisible by odd numbers <=5 ( that means the one of the numbers will be definitely divisible by (3, 5) say we have 100,102,104,106,108, We do have 100 divisible by 5 and (102 & 108) divisible by 3 Now say we have series of 7 consecutive odd number then one of them will be definitely divisible by odd numbers <=7( that means the one of the numbers will be definitely divisible by (3, 5, 7) say the series is 101, 103,105, 107,109 ,111, 113 So we have 105 is divisible by 7, 5& 3 and 109 is divisible by 3. We can expand from this result and get many more results that we can use during exams. Thanks to chetan2u




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