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If n is a positive integer, what is the remainder when n^2 is divided [#permalink]
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22 Apr 2016, 06:38
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If n is a positive integer, what is the remainder when n^2 is divided [#permalink]
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22 Apr 2016, 07:53
Stmnt 1> so n can be 1 ,4 ,9, 13 ,17 ,21 etc so n^2/4 will always leave a remainder of 1 Sufficient Stmnt 2 > n= 1,3,5,7,9 etc so n^2/4 will always leave a remainder of 1 Sufficient
Ans D



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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink]
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22 Apr 2016, 17:46
Statement 1: When n is divided by 4, the remainder is 1. so the Number is (4x + 1) On squaring this number and checking divisibility with 4 , we get R[(4x + 1)^2 / 4] = R[(16x^2 + 8x +1) /4] = 1 The remainder of the above equation is 1 regardless of value of x Sufficient
Statement 2: n is odd. Here the number can be (4x + 1) or (4x  1) As we saw above, the remainder will be 1 in the end Sufficient
Correct option: D



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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink]
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22 Apr 2016, 22:32
ronny123 wrote: Statement 1: When n is divided by 4, the remainder is 1. so the Number is (4x + 1) On squaring this number and checking divisibility with 4 , we get R[(4x + 1)^2 / 4] = R[(16x^2 + 8x +1) /4] = 1 The remainder of the above equation is 1 regardless of value of x Sufficient
Statement 2: n is odd. Here the number can be (4x + 1) or (4x  1) As we saw above, the remainder will be 1 in the end Sufficient
Correct option: D A is sufficient to answer this question.
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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink]
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22 Apr 2016, 23:31
Balajikarthick1990 wrote: ronny123 wrote: Statement 1: When n is divided by 4, the remainder is 1. so the Number is (4x + 1) On squaring this number and checking divisibility with 4 , we get R[(4x + 1)^2 / 4] = R[(16x^2 + 8x +1) /4] = 1 The remainder of the above equation is 1 regardless of value of x Sufficient
Statement 2: n is odd. Here the number can be (4x + 1) or (4x  1) As we saw above, the remainder will be 1 in the end Sufficient
Correct option: D A is sufficient to answer this question. Hi Balajikarthick1990, all prime number > 3 are of the form 6n +1 or 6n1.. Ofcourse vice versa need not be correct that is a number of 6n+1 or 6n1 form need not be a PRIME..so square of these two cases will be 1) (6n+1)^2 = 36n^2+12n+1..so 36n^2 and 12n are div by 4, irrespective of value of n.. out of 36n^2+12n+1, ONLY 1 is left so R=1 2) similarly (6n1)^2 = 36n^212n+1..so 36n^2 and 12n are div by 4, irrespective of value of n.. out of 36n^212n+1, ONLY 1 is left so R=1 statement II is also suff
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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink]
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22 Apr 2016, 23:36
chetan2u wrote: Balajikarthick1990 wrote: ronny123 wrote: Statement 1: When n is divided by 4, the remainder is 1. so the Number is (4x + 1) On squaring this number and checking divisibility with 4 , we get R[(4x + 1)^2 / 4] = R[(16x^2 + 8x +1) /4] = 1 The remainder of the above equation is 1 regardless of value of x Sufficient
Statement 2: n is odd. Here the number can be (4x + 1) or (4x  1) As we saw above, the remainder will be 1 in the end Sufficient
Correct option: D A is sufficient to answer this question. Hi Balajikarthick1990, all prime number > 3 are of the form 6n +1 or 6n1.. Ofcourse vice versa need not be correct that is a number of 6n+1 or 6n1 form need not be a PRIME..so square of these two cases will be 1) (6n+1)^2 = 36n^2+12n+1..so 36n^2 and 12n are div by 4, irrespective of value of n.. out of 36n^2+12n+1, ONLY 1 is left so R=1 2) similarly (6n1)^2 = 36n^212n+1..so 36n^2 and 12n are div by 4, irrespective of value of n.. out of 36n^212n+1, ONLY 1 is left so R=1 statement II is also suff Yes chetan sir...Misread..sorry...
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