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by okletsdothis » Fri Sep 03, 2010 6:30 am
If n is a prime number greater than 3, what is the
remainder when n2(n square) is divided by 12 ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 5


Ans is B

Can someone explain this to me in simpler language ?
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by Gurpinder » Fri Sep 03, 2010 6:44 am
okletsdothis wrote:If n is a prime number greater than 3, what is the
remainder when n2(n square) is divided by 12 ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 5


Ans is B

Can someone explain this to me in simpler language ?
N is a prime greater than 3. So N = 5. N^2 = 25. 25/12 = 24 with a remainder 1. So (B)
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by okletsdothis » Fri Sep 03, 2010 6:50 am
YOu have considered only 1 prime number greater than 3. cant be true for all prime numbers ?
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by Gurpinder » Fri Sep 03, 2010 6:57 am
okletsdothis wrote:YOu have considered only 1 prime number greater than 3. cant be true for all prime numbers ?
Good thinking! but this is not a data sufficiency problem. If we start considering other prime numbers, the list will go on for ever...
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by debmalya_dutta » Fri Sep 03, 2010 7:10 am
okletsdothis wrote:YOu have considered only 1 prime number greater than 3. cant be true for all prime numbers ?
You can try it out for any number of prime numbers
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by zino0067 » Fri Sep 03, 2010 8:04 am
okletsdothis wrote:If n is a prime number greater than 3, what is the
remainder when n2(n square) is divided by 12 ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 5


Ans is B

Can someone explain this to me in simpler language ?
n^2-1 (mod 12)≡(n+1)(n-1) (mod 12)
Now n is odd number.
n^2-1 (mod 12)≡0
n^2 (mod 12)≡1

Therefore Ans is B.
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by GMATGuruNY » Fri Sep 03, 2010 12:07 pm
okletsdothis wrote:YOu have considered only 1 prime number greater than 3. cant be true for all prime numbers ?
If different prime numbers could yield different remainders, there would be no correct answer choice.
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