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If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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25 May 2012, 02:00
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If n is a positive integer, is n^2  1 divisible by 24? (1) n is a prime number. (2) n is greater than 191
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If n is a positive integer, is n^2  1 divisible by 24?(1) n is a prime number > if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient. (2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer). (1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^21=(n1)(n+1) > out of three consecutive integers (n1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n1) or (n+1), so (n1)(n+1) is divisible by 3. Next, since n is odd then (n1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n1)(n+1) is divisible by 2*4=8. We have that (n1)(n+1) is divisible by both 3 and 8 so (n1)(n+1) is divisible by 3*8=24. Sufficient. Answer: C. Hope it's clear.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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27 May 2012, 01:59
Bunuel wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number > if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.
(2) n is greater than 191. Clearly insufficient (consider n=24 for a NO answer and n=17 for an YES answer).
(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^21=(n1)(n+1) > out of three consecutive integers (n1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n1) or (n+1), so (n1)(n+1) is divisible by 3. Next, since n is odd then (n1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n1)(n+1) is divisible by 2*4=8. We have that (n1)(n+1) is divisible by both 3 and 8 so (n1)(n+1) is divisible by 3*8=24. Sufficient.
Answer: C.
Hope it's clear. bunuel can you please explain the colored part of your answer. if n is not there, why it n1 & n+1 must be divisible by 3. it may be possible that only n is divisible by 3 example; n1=2,n=3,n+1=4 i.e 2,3,4 if 3 is not there,, why 2x4 must be divisible by 3 ??
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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27 May 2012, 04:38
321kumarsushant wrote: Bunuel wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number > if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.
(2) n is greater than 191. Clearly insufficient (consider n=24 for a NO answer and n=17 for an YES answer).
(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^21=(n1)(n+1) > out of three consecutive integers (n1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n1) or (n+1), so (n1)(n+1) is divisible by 3. Next, since n is odd then (n1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n1)(n+1) is divisible by 2*4=8. We have that (n1)(n+1) is divisible by both 3 and 8 so (n1)(n+1) is divisible by 3*8=24. Sufficient.
Answer: C.
Hope it's clear. bunuel can you please explain the colored part of your answer. if n is not there, why it n1 & n+1 must be divisible by 3. it may be possible that only n is divisible by 3 example; n1=2,n=3,n+1=4 i.e 2,3,4 if 3 is not there,, why 2x4 must be divisible by 3 ?? For (1)+(2) we have that n is odd and not a multiple of 3. Next, (n1), n and n+1 represent three consecutive integers. Out of ANY three consecutive integers one is always divisible by 3, we know that it's not n, so it must be either n1 or n+1. Hope it's clear.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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29 May 2012, 07:16
Hi Bunuel... n is greater than 191. Clearly insufficient (consider n=24 for a NO answer and n=17 for an YES answer).....I did not understand when n> 191, why are we considering n =24,17 <191?



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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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29 May 2012, 09:56



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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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I have another solution which came to me. Clearly first and second statement alone each can not be sufficient to find out whether expression is divisible by 24 or not. Now considering both, we know that n is prime number which is greater than 191. All prime numbers except 2 and 3 are in the form of 6n+1 or 6n1.
n^2 1 = (6n+1)^2  1 or n^2 1 = (6n  1)^2  1
n^2 1 = 36n^2 + 12n or n^2 1 = 36n^2  12n
= 12(n^2 + n) or 12(n^2  n)
Now we know that this expression is divisible by 12 and this will be divisible by 24 if (n^2+n) or (n^2n) is even numbers. As we know that n is prime number which is greater than 191 so it has to be odd so n^2 will also be odd and now we can say that n^2+n will be even (odd+odd = even). Same way n^2  n will be even.
So this expression will definitely divisible by 24.
Bunuel  Please let me know if I am doing any wrong over here.



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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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sandal85 wrote: I have another solution which came to me. Clearly first and second statement alone each can not be sufficient to find out whether expression is divisible by 24 or not. Now considering both, we know that n is prime number which is greater than 191. All prime numbers except 2 and 3 are in the form of 6n+1 or 6n1.
n^2 1 = (6n+1)^2  1 or n^2 1 = (6n  1)^2  1
n^2 1 = 36n^2 + 12n or n^2 1 = 36n^2  12n
= 12(n^2 + n) or 12(n^2  n)
Now we know that this expression is divisible by 12 and this will be divisible by 24 if (n^2+n) or (n^2n) is even numbers. As we know that n is prime number which is greater than 191 so it has to be odd so n^2 will also be odd and now we can say that n^2+n will be even (odd+odd = even). Same way n^2  n will be even.
So this expression will definitely divisible by 24.
Bunuel  Please let me know if I am doing any wrong over here. Your approach is correct, but when expressing prime number \(n\), you shouldn't use the same variable (\(n\)). Also there are some other little mistakes. It should be: since any prime number greater than 3 could be expressed as \(6k+1\) or\(6k+5\) (\(6k1\)), where \(k\) is an integer >1, then \(n=6k+1\) or \(n=6k1\). \(n^21=36k^2+12k=12k(3k+1)\) or \(n^21=36k^212k=12k(3k1)\). We can see that \(n^21\) is divisible by 12. Next, notice that if \(k=odd\) then \(3k+1=even\) (\(3k1=even\)) and if \(k=even\) then \(3k+1=odd\) (\(3k1=odd\)). So, \(n^21\) is also divisible by 2, which means that \(n^21\) is divisible by 12*2=24. Hope it's clear.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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31 May 2012, 04:45
321kumarsushant wrote: bunuel can you please explain the colored part of your answer. if n is not there, why it n1 & n+1 must be divisible by 3. it may be possible that only n is divisible by 3 example; n1=2,n=3,n+1=4 i.e 2,3,4 if 3 is not there,, why 2x4 must be divisible by 3 ?? The assumption is that n is a prime and is greater than 191 which is the reason why N is not a multiple of 3. If you factorize the given equation it comes to (n+1)(n1) and which when combine with N comes to a product of three consecutive numbers. and it is know that a product of three consecutive numbers are divisible by 3. hence one of the numbers is divisible by 3. I hope this clarifies your reasoning.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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kunalbh19 wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number. (2) n is greater than 191 St 1 : If n =2 then n^21 is not divisibly by 24 but if n=7 then n^21 is divisible by 24 There is a there property that for any prime (p) greater than 6, p^2 1 is always divisible by 6 Option A,D ruled out St 2 says n>191 again if n is not prime than n^21 may or may not be divisible by 24 but if n is prime and greater than 191 then it surely divisible by 24 since 2 choices are possible so option B is ruled out Combining we get n is prime and n>191 and hence the expression n^21 will be divisible by 24 for any value of n Ans is C
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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kunalbh19 wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number. (2) n is greater than 191 Any prime number greater than 3 can be represented as \(6k\pm1\) but NOT necessarily viceversa, where k = 1,2,3..etc I. If n is prime,\(n^21 = (6k+1)^21\) > (6k+2)*6k > 12*k*(3k+1) . Either k is odd and (3k+1) is even OR k is even and (3k+1) is odd. Either ways, with one even factor in the expression, \(n^21\) would always have a factor of 24. II.If n is prime, \(n^21 = (6k1)^21\) > (6k2)*6k > 12*k*(3k1) . Either k is odd and (3k1) is even OR k is even and (3k1) is odd. Either ways, with one even factor in the expression, \(n^21\) would always have a factor of 24. Thus, any prime number n>3,\(n^21\) is ALWAYS divisible by 24. F.S 1 We don't know whether n>3 or not. Insufficient. F.S 2 We don't know whether n is prime or not. For n= 240 we get a YES, for n = 241, we get a NO.Insufficient. Taking both togehter, we know that n is prime and n>3. Thus, \(n^21\) will always be divisible by 24. Sufficient. C.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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kunalbh19 wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number. (2) n is greater than 191 Question: YES or NO type (1) n is a prime number. n can be 2 or 3 or >=5 both bearing YES and NO respectively =>Not sufficient (2) n is greater than 191 Lets take a no say 200 200^2 = 40000 40000 1 = 3999 = 3*1333 we need another 8 in the factors to make it divisible by 24 , Not possible , although for all the primary no >=5 the question is true, Thus, for all primary no's >191, it will hold true. Again a YES and a NO =>Not sufficient. (1)+(2) again, we already considered this argument, this will definitely result in a YES =>Sufficient Ans: C
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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23 Jul 2013, 21:47
Bunuel wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number > if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.
(2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer).
(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^21=(n1)(n+1) > out of three consecutive integers (n1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n1) or (n+1), so (n1)(n+1) is divisible by 3. Next, since n is odd then (n1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n1)(n+1) is divisible by 2*4=8. We have that (n1)(n+1) is divisible by both 3 and 8 so (n1)(n+1) is divisible by 3*8=24. Sufficient.
Answer: C.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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This question can be rephrased as follows: Is n^2  1 = 24k ? Is n^2 = 24k+1? (24 k + 1 is odd) so question is: Is n^2 = odd? (since n is positive integer, and if n^2 = odd then n must be odd) so question is : Is n = odd ? statement 1: n is prime : If n = 2 = even, if n = 3 = odd .Not suff Statement 2: n > 191 , n = 192 = even, n = 193 (odd) . Not suff 1 + 2 : n is prime and > 191 = n is odd. Sufficient C
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If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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01 Aug 2014, 23:45
Bunuel wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number > if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.
(2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer).
(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^21=(n1)(n+1) > out of three consecutive integers (n1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n1) or (n+1), so (n1)(n+1) is divisible by 3. Next, since n is odd then (n1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n1)(n+1) is divisible by 2*4=8. We have that (n1)(n+1) is divisible by both 3 and 8 so (n1)(n+1) is divisible by 3*8=24. Sufficient.
Answer: C.
Hope it's clear. they are asking us whether (n+1)(n1) is divisible by 24 if n is a prime number greater than 191 then n is odd ==> (n+1) and (n1) are even integers out of n, (n+1), and (n1) one number must be divisible by 3 if n is divisible by 3 then (n+1) and (n1) are not divisible by 3 > Insufficient if n is not divisible by 3 then its either (n+1) or (n1) which is divisible by 3 > Sufficient So answer must be E Bunuel, correct me if i am wrong!



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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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02 Aug 2014, 02:06
AkshayDavid wrote: Bunuel wrote: If n is a positive integer, is n^2  1 divisible by 24?
(1) n is a prime number > if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.
(2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer).
(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^21=(n1)(n+1) > out of three consecutive integers (n1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n1) or (n+1), so (n1)(n+1) is divisible by 3. Next, since n is odd then (n1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n1)(n+1) is divisible by 2*4=8. We have that (n1)(n+1) is divisible by both 3 and 8 so (n1)(n+1) is divisible by 3*8=24. Sufficient.
Answer: C.
Hope it's clear. they are asking us whether (n+1)(n1) is divisible by 24 if n is a prime number greater than 191 then n is odd ==> (n+1) and (n1) are even integers out of n, (n+1), and (n1) one number must be divisible by 3 if n is divisible by 3 then (n+1) and (n1) are not divisible by 3 > Insufficient if n is not divisible by 3 then its either (n+1) or (n1) which is divisible by 3 > Sufficient So answer must be E Bunuel, correct me if i am wrong! The correct answer is C (check the spoiler in the original post), so you must be wrong somewhere. n is a prime greater than 191, hence it CANNOT be divisible by 3.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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12 Oct 2017, 20:51
Hi Bunuel,
I have a question, "How did you figure out that THE PRIME NUMBERS >191 MUST BE 'ODD' AND 'NOT DIVISIBLE BY 3'?" I thought, prime numbers do end in 1,3, 7 or 9 and are odd (except "2")! Is what your statement a part of the definition for prime numbers and can be used for all numbers(not just those >191) or not?
Kind regards, Sep.



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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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12 Oct 2017, 20:57
sepehr23 wrote: Hi Bunuel,
I have a question, "How did you figure out that THE PRIME NUMBERS >191 MUST BE 'ODD' AND 'NOT DIVISIBLE BY 3'?" I thought, prime numbers do end in 1,3, 7 or 9 and are odd (except "2")! Is what your statement a part of the definition for prime numbers and can be used for all numbers(not just those >191) or not?
Kind regards, Sep. All primes except 2 are odd. Only prime that is divisible by 3 is 3 itself. So, if n is a prime number greater than 191, then n is odd and not a multiple of 3.
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Re: If n is a positive integer, is n^2  1 divisible by 24? [#permalink]
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Great ! Thanks for the quick answer. Consequently, the final result is that "The combined answer is sufficient to say that 'n^2  1' is NOT divisible by 24, right?"




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