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# If n is an integer, what is the remainder when (n – 3)^2 is divided by

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Math Expert
Joined: 02 Sep 2009
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If n is an integer, what is the remainder when (n – 3)^2 is divided by  [#permalink]

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27 Jan 2020, 02:04
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45% (medium)

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62% (01:32) correct 38% (01:47) wrong based on 13 sessions

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If n is an integer, what is the remainder when (n – 3)^2 is divided by 4?

(1) n is divisible by 2.
(2) n is divisible by 4.

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Re: If n is an integer, what is the remainder when (n – 3)^2 is divided by  [#permalink]

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27 Jan 2020, 02:19
IMO Option D

Both statements are alone sufficient to answee the question.

Any multiple of 2 or 4 gives remainder 1 when divided by(n-3)^2

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Re: If n is an integer, what is the remainder when (n – 3)^2 is divided by  [#permalink]

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27 Jan 2020, 02:24
1
If n is an integer, what is the remainder when $$(n – 3)^2$$ is divided by 4?

(Statement1): n is divisible by 2.
-->let's say that n =2k (k -integer)
$$(n – 3)^{2}= (2k-3)^{2}= 4k^{2} -12k+ 9 --> 4(k^{2} -3k+ 2)+ 1$$
--> the remainder is 1.
Sufficient

(Statement2): n is divisible by 4.
-->let's say that n =4k (k -integer)
$$(n – 3)^{2}= (4k-3)^{2}= 16k^{2} -24k+ 9 --> 4(4k^{2} -8k+ 2)+ 1$$
--> the remainder is 1.
Sufficient

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Re: If n is an integer, what is the remainder when (n – 3)^2 is divided by  [#permalink]

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27 Jan 2020, 02:51
Bunuel wrote:
If n is an integer, what is the remainder when (n – 3)^2 is divided by 4?

(1) n is divisible by 2.
(2) n is divisible by 4.

Question: remainder when (n – 3)^2 is divided by 4?

Statement 1:n is divisible by 2.

@n=4, Remainder[(4-3)^2 / 4] = 1
@n=6, Remainder[(6-3)^2 / 4] = 1
@n=4, Remainder[(8-3)^2 / 4] = 1

i.e. Reminder is always 1 hence

SUFFICIENT

STatement 2: n is divisible by 4

As checked in statement 1, the remainder is always 1 when n is even hence when n is a multiple of 4 even then the remainder will be 1

SUFFICIENT

P.S. Industio n method is one of the most promising method when algebraic simplification is lengthy and/or complex
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Re: If n is an integer, what is the remainder when (n – 3)^2 is divided by   [#permalink] 27 Jan 2020, 02:51
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