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27 Jan 2020, 03:04
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D
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Difficulty:
45%
(medium)
Question Stats:
63% (01:42) correct
37%
(01:45)
wrong
based on 111
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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.
(1) n is divisible by 2.
(2) n is divisible by 4.
27 Jan 2020, 03:19
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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
Posted from my mobile device
Both statements are alone sufficient to answee the question.
Any multiple of 2 or 4 gives remainder 1 when divided by(n-3)^2
Posted from my mobile device
27 Jan 2020, 03:24
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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
The answer is D.
(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
The answer is D.
