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If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is

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Bunuel
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If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is [#permalink] New post   15 Mar 2019, 00:26
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If n is a positive integer, when (2n + 2)^2 is divided by 4 the remainder is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
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A
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Re: If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is [#permalink] New post   15 Mar 2019, 02:34
Bunuel wrote:
If n is a positive integer, when (2n + 2)^2 is divided by 4 the remainder is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Note: n is a positive integer.

(2n + 2) = even + even= even.

Set a minimum value.

n = 1.

2*1 + 2 = 4. when 4 is divided by 4 remainder is 0. It will be true for all cases.

The best answer s A .
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Re: If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is [#permalink] New post   15 Mar 2019, 02:53
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Expand (2n+2)^2
4n^2 + 8n + 4... Each term is divisible by 4... Remainder will always be 0.

Alternatively plugging in Will also work..

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Re: If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is [#permalink] New post   16 Mar 2019, 02:45
Bunuel wrote:
If n is a positive integer, when (2n + 2)^2 is divided by 4 the remainder is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


for all values of n , in expression(2n + 2)^2
we get remainder as 0 when divided by 4
IMO A
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Re: If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is [#permalink] New post   17 Mar 2019, 01:55
(2n+2)^2 = 2^2 (n+1)^2 => 4 (n+1)^2

4(n+1)^2 divided by 4 will be (n+1)^2, hence remainder is 0
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Re: If n is an integer, when (2n + 2)^2 is divided by 4 the remainder is [#permalink]
17 Mar 2019, 01:55
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