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If n is a positive integer, what is the remainder when n^2 is divided

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Bunuel
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If n is a positive integer, what is the remainder when n^2 is divided [#permalink] New post   22 Apr 2016, 06:38
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If n is a positive integer, what is the remainder when n^2 is divided by 4?

(1) When n is divided by 4, the remainder is 1.
(2) n is odd.
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D
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ronny123
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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink] New post   22 Apr 2016, 17:46
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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] New post   22 Apr 2016, 23:31
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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 6n-1..
Ofcourse vice versa need not be correct that is a number of 6n+1 or 6n-1 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 (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

statement II is also suff
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If n is a positive integer, what is the remainder when n^2 is divided [#permalink] New post   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] New post   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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Balajikarthick1990
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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink] New post   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 6n-1..
Ofcourse vice versa need not be correct that is a number of 6n+1 or 6n-1 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 (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

statement II is also suff



Yes chetan sir...Misread..sorry...
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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink] New post   01 Aug 2020, 23:30
Bunuel wrote:
If n is a positive integer, what is the remainder when n^2 is divided by 4?

(1) When n is divided by 4, the remainder is 1.
(2) n is odd.


Asked: If n is a positive integer, what is the remainder when n^2 is divided by 4?

(1) When n is divided by 4, the remainder is 1.
n = 4k + 1
n^2 = 16k^2 + 8k + 1
The remainder when n^2 is divided by 4 = 1
SUFFICIENT

(2) n is odd.
n = 2k + 1
n^2 = 4k^2 + 4k + 1
The remainder when n^2 is divided by 4 = 1
SUFFICIENT

IMO D
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Re: If n is a positive integer, what is the remainder when n^2 is divided [#permalink] New post   10 Jul 2022, 02:21
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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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