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If n is a positive integer greater than 6, what is the remai
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Updated on: 19 May 2014, 02:25
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If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3. (2) n^2 – 1 is even.
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Originally posted by russ9 on 18 May 2014, 11:58.
Last edited by Bunuel on 19 May 2014, 02:25, edited 1 time in total.
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Re: If n is a positive integer greater than 6, what is the remai
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19 May 2014, 02:34
If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3 > (n1)(n+1) is not divisible by 3 > neither n1 not n+1 is divisible by 3. Since from any 3 consecutive integers one is divisible by 3, then n must be divisible by 3. If n is even too (12, 18, ...) then the remainder is 0 but if n is odd (9, 15, ...) then the reminder is 3. Not sufficient. (2) n^2 – 1 is even > n^2 = even + 1 = odd > n=odd. Not sufficient. (1)+(2) n is odd multiple of 3: 9, 15, 21, ... The remainder when n is divided by 6 is 3. Sufficient. Answer: C.
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Re: If n is a positive integer greater than 6, what?
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18 May 2014, 19:01
How it can be C, because the n value where n^21 is not divisible by 6 is 9; and also if we consider n^21 is even we can get a lot numbers 48, 120, 80 etc.



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Re: If n is a positive integer greater than 6, what?
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18 May 2014, 19:53
krishna789 wrote: How it can be C, because the n value where n^21 is not divisible by 6 is 9; and also if we consider n^21 is even we can get a lot numbers 48, 120, 80 etc. Q:If n is a positive integer greater than 6, what is the remainder when n is divided by 6?
(1) n2 – 1 is not divisible by 3.
(2) n2 – 1 is even. Ans is C and here is how St 1 tells us that n^21 is not divisible by 3 and we know n>6 so possible values of n =9,12,15,18...... Now if n =12,18,24... then n/6 gives remainder as 0 but if n=9,15,21 then remainder is 3 So from St1 we see that remainder will follow a pattern of 0,1,0,1.... Not sufficient from St 2 we we have n^21=even or n^2= Even +1 =odd, so we see that n is odd. Now if n=7 then n/6 gives remainder 1 but if n=9 then remainder is 3
if n=11 then remainder is 5 So St2 is not sufficient alone Combining we see that n is odd and n>6 so possible value of n satisfying the above conditions are n=9,15,21,27.....so on Remainder will be 3 Ans is C
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Re: If n is a positive integer greater than 6, what?
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Updated on: 19 May 2014, 14:12
russ9 wrote: If N is a positive integer greater than 6, what is the remainder when N is divided by 6? (1) N^2 – 1 is not divisible by 3. (2) N^2 – 1 is even. N is a positive integer > 6, what is the remainder when N is divided by 6?(1) N^2 – 1 is not divisible by 3.> N^2  1 = (N1)*(N+1) , and this is not divisible by three. Notice that (N1)*N*(N+1) is the product of three consecutive positive integers. Such a product will always contain exactly one multiple of 3. If this multiple of three is not (N1) or (N+1), then the multiple of 3 must be N. If N = 3*3=9, the remainder when divided by 6 is 3. If N = 3*4=12, the remainder when divided by 6 is 0. > Not sufficient. (Note: The product of K consecutive positive integers is always divisible by K!) (2) N^2 – 1 is even> N^2 = even + 1 = odd > N is odd. If N = 7, the remainder when divided by 6 is 1. If N = 9, the remainder when divided by 6 is 3. > Not sufficient. (1) and (2)N is an odd multiple of 3 greater than 6. > N = 3*(2k1) = 6k  3 for k >1. Thus, since N is three less than a multiple of six, the remainder is always 3. Sufficient.
Originally posted by Reinfrank2011 on 18 May 2014, 21:40.
Last edited by Reinfrank2011 on 19 May 2014, 14:12, edited 1 time in total.



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Re: If n is a positive integer greater than 6, what is the remai
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19 May 2014, 16:49
Bunuel wrote: If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3 > (n1)(n+1) is not divisible by 3 > neither n1 not n+1 is divisible by 3. Since from any 3 consecutive integers one is divisible by 3, then n must be divisible by 3. If n is even too (12, 18, ...) then the remainder is 0 but if n is odd (9, 15, ...) then the reminder is 3. Not sufficient. (2) n^2 – 1 is even > n^2 = even + 1 = odd > n=odd. Not sufficient.
(1)+(2) n is odd multiple of 3: 9, 15, 21, ... The remainder when n is divided by 6 is 3. Sufficient.
Answer: C. Thanks for the explanation. I can definitely follow what you have outlined above but I wouldn't have been able to derive it on my own. Two questions: 1) Is it better to plug in numbers for these type of problems or work through the theory? 2) How do I get better at these specific types of problems? Is it just practice and if so, can you suggest similar problems please? Thanks!



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Re: If n is a positive integer greater than 6, what is the remai
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20 May 2014, 00:38
russ9 wrote: Bunuel wrote: If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3 > (n1)(n+1) is not divisible by 3 > neither n1 not n+1 is divisible by 3. Since from any 3 consecutive integers one is divisible by 3, then n must be divisible by 3. If n is even too (12, 18, ...) then the remainder is 0 but if n is odd (9, 15, ...) then the reminder is 3. Not sufficient. (2) n^2 – 1 is even > n^2 = even + 1 = odd > n=odd. Not sufficient.
(1)+(2) n is odd multiple of 3: 9, 15, 21, ... The remainder when n is divided by 6 is 3. Sufficient.
Answer: C. Thanks for the explanation. I can definitely follow what you have outlined above but I wouldn't have been able to derive it on my own. Two questions: 1) Is it better to plug in numbers for these type of problems or work through the theory? 2) How do I get better at these specific types of problems? Is it just practice and if so, can you suggest similar problems please? Thanks! 1. Questions on remainders are good for plugin, but not all of them. It depends on a question and your skills/preferences what approach to choose. 2. By studying theory and practicing: Hope this helps.
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Re: If n is a positive integer greater than 6, what is the remai
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14 Apr 2017, 19:11
Here is my take! Statement 1: n^2  1 is not divisible by 3 > (n1)(n+1) is not divisible by 3 > n is divisible by 3 > insuff Statement 2: n^2  1 is odd > (n1)(n+1) is odd > n 1 is even > n is odd > insuff Statement 1+2: n is divisible by 3 > n = 3a n is even > n = 2b +1 So, n = 6c +3 > remainder when n is divided by 6 is 3 Hence, C!
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Re: If n is a positive integer greater than 6, what is the remai
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22 May 2017, 19:44
russ9 wrote: If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3. (2) n^2 – 1 is even. Goal: We need to know the remainder when n is divided by 6. Statement 1: (N^21)/3 =int. So (n+1)(n1)/3 is not an integer. Thus, n must be a multiple of three because every three consecutive integers has at least one multiple of three within it. Not sufficient because all it tells us is that n is a multiple of 3, so its remainder when divided by 6 has multiple potential values. Statement 2: N^2  1 is even. Hence (n+1)*(n1) is even, so n must be odd. Not sufficient because some odd numbers have different remainders when divided by 6. Combined: We know that N is an integer greater than 6 that is odd and a multiple of 3. Start testing cases: n=9 (r=3); n=15 (r=3); and so on. Sufficient.



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If n is a positive integer greater than 6, what is the remai
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01 Feb 2020, 09:07
russ9 wrote: If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3. (2) n^2 – 1 is even. The problem can be solved algebraically or using smart numbers (testing cases). Both methods are shown below. (1) NOT SUFFICIENT: Algebra The expression n^\(2\) – 1 = (n – 1)(n + 1). Together with n, the three expressions represent three consecutive integers: n – 1, n, n + 1. Statement 2 indicates that the product (n – 1)(n + 1) is not divisible by the prime number 3; therefore, neither n – 1 nor n + 1 is a multiple of 3. Since every third integer is a multiple of 3, it follows that n must be a multiple of 3. This fact alone is not sufficient, though, since different multiples of 3 can give different remainders upon division by 6. For example, 9/6 gives a remainder of 3, and 12/6 gives a remainder of 0. Testing Cases Find values of n > 6 that satisfy statement 1. The value n = 7 does not satisfy the statement, since 72 – 1 = 48 is divisible by 3. The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is divisible by 3. The value n = 9 satisfies the statement, since 92 – 1 = 80 is not divisible by 3. The remainder of 9/6 is 3. The value n = 10 does not satisfy the statement, since 102 – 1 = 99 is divisible by 3. The value n = 11 does not satisfy the statement, since 112 – 1 = 120 is divisible by 3. The value n = 12 satisfies the statement, since 122 – 1 = 143 is not divisible by 3. The remainder of 12/6 is 0. At least two different remainders are possible, so the statement is insufficient. (2) NOT SUFFICIENT: Algebra If n\(2\) – 1 is even, then \(n2\) must be odd, so n itself is odd. This fact alone is not sufficient, though, since different odd numbers can give different remainders upon division by 6. For example, 7/6 gives a remainder of 1, and 9/6 gives a remainder of 3. Testing Cases Find values of n > 6 that satisfy the statement. The value n = 7 satisfies the statement, since 72 – 1 = 48 is even. The remainder of 7/6 is 1. The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is not even. The value n = 9 satisfies the statement, since 92 – 1 = 80 is even. The remainder of 9/6 is 3. At least two different remainders are possible, so the statement is insufficient. (1) AND (2) SUFFICIENT: Statement (1) indicates that n is a multiple of 3 and statement (2) indicates that n is odd. From this point, you can either use algebra or test numbers. Algebra According to the two statements, n must be 3 times some odd integer, so it can be written as n = 3(2k + 1), where k is an integer. Distributing gives n = 6k + 3, which is 3 more than a multiple of 6. Therefore, the remainder upon dividing n by six must be 3. Testing Cases Test numbers that are multiples of 3, odd, and greater than 6: If n = 9, then 9 / 6 = 1 remainder 3. If n = 15, then 15 / 6 = 2 remainder 3. If n = 21, then 21 / 6 = 3 remainder 3



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Re: If n is a positive integer greater than 6, what is the remai
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01 Feb 2020, 22:03
russ9 wrote: If n is a positive integer greater than 6, what is the remainder when n is divided by 6? (1) n^2 – 1 is not divisible by 3. (2) n^2 – 1 is even. The problem can be solved algebraically or using smart numbers (testing cases). Both methods are shown below. (1) NOT SUFFICIENT: Algebra The expression n2 – 1 = (n – 1)(n + 1). Together with n, the three expressions represent three consecutive integers: n – 1, n, n + 1. Statement 2 indicates that the product (n – 1)(n + 1) is not divisible by the prime number 3; therefore, neither n – 1 nor n + 1 is a multiple of 3. Since every third integer is a multiple of 3, it follows that n must be a multiple of 3. This fact alone is not sufficient, though, since different multiples of 3 can give different remainders upon division by 6. For example, 9/6 gives a remainder of 3, and 12/6 gives a remainder of 0. Testing Cases Find values of n > 6 that satisfy statement 1. The value n = 7 does not satisfy the statement, since 72 – 1 = 48 is divisible by 3. The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is divisible by 3. The value n = 9 satisfies the statement, since 92 – 1 = 80 is not divisible by 3. The remainder of 9/6 is 3. The value n = 10 does not satisfy the statement, since 102 – 1 = 99 is divisible by 3. The value n = 11 does not satisfy the statement, since 112 – 1 = 120 is divisible by 3. The value n = 12 satisfies the statement, since 122 – 1 = 143 is not divisible by 3. The remainder of 12/6 is 0. At least two different remainders are possible, so the statement is insufficient. (2) NOT SUFFICIENT: Algebra If n2 – 1 is even, then n2 must be odd, so n itself is odd. This fact alone is not sufficient, though, since different odd numbers can give different remainders upon division by 6. For example, 7/6 gives a remainder of 1, and 9/6 gives a remainder of 3. Testing Cases Find values of n > 6 that satisfy the statement. The value n = 7 satisfies the statement, since 72 – 1 = 48 is even. The remainder of 7/6 is 1. The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is not even. The value n = 9 satisfies the statement, since 92 – 1 = 80 is even. The remainder of 9/6 is 3. At least two different remainders are possible, so the statement is insufficient. (1) AND (2) SUFFICIENT: Statement (1) indicates that n is a multiple of 3 and statement (2) indicates that n is odd. From this point, you can either use algebra or test numbers. Algebra According to the two statements, n must be 3 times some odd integer, so it can be written as n = 3(2k + 1), where k is an integer. Distributing gives n = 6k + 3, which is 3 more than a multiple of 6. Therefore, the remainder upon dividing n by six must be 3. Testing Cases Test numbers that are multiples of 3, odd, and greater than 6: If n = 9, then 9 / 6 = 1 remainder 3. If n = 15, then 15 / 6 = 2 remainder 3. If n = 21, then 21 / 6 = 3 remainder 3. Will this always return a remainder of 3? Yes! An odd multiple of 3 will always have a remainder when divided by 6 (because 6 is even). Further, an even multiple of 3 will always have a remainder of zero (because an even multiple of 3 is divisible by both 2 and 3). Therefore, the odd multiple of 3, which is always 3 more than an even multiple of 3, will always have a remainder of 3. The correct answer is (C).




Re: If n is a positive integer greater than 6, what is the remai
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