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M27-07

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M27-07  [#permalink]

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New post 16 Sep 2014, 00:27
2
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

68% (00:55) correct 32% (00:59) wrong based on 199 sessions

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Re M27-07  [#permalink]

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New post 26 Apr 2016, 23:54
1
Official Solution:


(1) \(p\) is greater than 3. Clearly insufficient: different values of \(p\) will give different values of the remainder.

(2) \(p\) is a prime. Also insufficient: if \(p=2\) then the remainder is 4 but if \(p=3\) then the remainder is 9.

(1)+(2) You can proceed with number plugging and try several prime numbers greater than 3 to see that the remainder will always be 1 (for example try \(p=5\), \(p=7\), \(p=11\)).

If you want to double-check this with algebra, you should apply the following property of the prime number: any prime number greater than 3 can be expressed either as \(p=6n+1\) or \(p=6n-1\).

If \(p=6n+1\), then \(p^2=36n^2+12n+1\) which gives remainder 1 when divided by 12;

If \(p=6n-1\), then \(p^2=36n^2-12n+1\) which also gives remainder 1 when divided by 12.


Answer: C
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Re: M27-07  [#permalink]

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New post 10 Jul 2018, 06:12
chetan2u niks18 pushpitkc Abhishek009

I was stumped a bit on this question by trusting my gut that
I have infinite prime numbers greater than 3 and did not bother
to square them to check for remainder after dividing by 12.

Could prime factorization of 12 would have been of any use?
I did not understand algebraic approach used by Bunuel
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Re: M27-07  [#permalink]

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New post 10 Jul 2018, 06:32
adkikani wrote:
chetan2u niks18 pushpitkc Abhishek009

I was stumped a bit on this question by trusting my gut that
I have infinite prime numbers greater than 3 and did not bother
to square them to check for remainder after dividing by 12.

Could prime factorization of 12 would have been of any use?
I did not understand algebraic approach used by Bunuel


The prime numbers greater than 3 are of the form 6k+-1...
Why..
Because...
6k is div by 2 and 3...
6k+1 not by 2 and 3.. chances of it to be prime
6k+2 by 2
6k +3 by 3
6k+4 by 2
6k+5 not by 2 and 3, so may be prime again but 6k+5=6k+6-1=6(k+1)-1
6k+6 is again of the form 6(k+1)

So only possibility is 6k+1 or 6k-1
When you square them (6k+1)=^2=36k^2+12k+1...
All terms have 12 except 1... So remainder is 1

OR
Just pick up some primes
5...5^3=25=12*2+1.. R is 1
7..7^2=49=12*4+1...R is 1
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: M27-07  [#permalink]

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New post 17 Sep 2018, 07:40
If p is a positive integer
statement 1: p>3, then P2 can be 16, 17, 18 etc and the remainder can be many numbers. hence insufficient.
Statement 2:P is prime, P2 can be 9,25,49 etc and hence remainder can be any number.
let us combine 1 and 2 statements, then P2 can be 25,49,121 etc and the remainder can be 1 only. hence answer is C
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Re: M27-07  [#permalink]

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New post 17 Sep 2018, 08:43
1) if p is 4 then remainder of \(p^2\) is 4, but if p is 5 then remainder is 1. INSUFFICIENT

2) if p is 2, remainder is 4, but is p is 3 then remainder is 3. INSUFFICIENT

Now by combining both statements, P should be greater than 3 and is a prime number then in all the possible cases, remainder of \(p^2\) is always 1.
So Answer is C
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Re: M27-07  [#permalink]

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New post 23 Sep 2018, 08:32
Bunuel wrote:
If \(p\) is a positive integer, what is the remainder when \(p^2\) is divided by 12?


(1) \(p\) is greater than 3.

(2) \(p\) is a prime.


Made a silly mistake by not testing 2 or 3 for Statement #2.
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Re: M27-07  [#permalink]

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New post 30 Sep 2018, 04:54
I messed up here by not continuing with the calculation. Because there are many prime numbers that are greater than 3, I assumed the remainder for them all would be different when divided by 12 but it is not so. 5, 7, 11, 13 and 59 all yield a remainder of 1. Keeping that in mind, I assume, it will be the same for any other numbers too.
And here I thought it was an easy question!
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Re: M27-07 &nbs [#permalink] 30 Sep 2018, 04:54
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