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# If p is a positive integer, what is a remainder when p^2 is

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If p is a positive integer, what is a remainder when p^2 is [#permalink]

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07 Oct 2009, 18:37
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This is a DS problem I made up with help of my old math school book.

If p is a positive integer, what is a remainder when p^2 is divided by 12?

(1) p>3.
(2) p is a prime.

Please tell me how hard is it? I marked it as 700+ but many of you might find it easy. Anyway GMAT offers lot of such kind of problems and this one will be good for practice.

Fill free to comment. Thanks.
[Reveal] Spoiler: OA

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07 Oct 2009, 18:59
Bunuel wrote:
This is a DS problem I made up with help of my old math school book.

If p is a positive integer, what is a remainder when p^2 is divided by 12?

(1) p>3.
(2) p is a prime.

Please tell me how hard is it? I marked it as 700+ but many of you might find it easy. Anyway GMAT offers lot of such kind of problems and this one will be good for practice.

Fill free to comment. Thanks.

Answer is C according to me.
The only thing I am not able to understand is, why always remainder is 1 in this case. Good question though.
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07 Oct 2009, 19:16
I guess you did number plugging?

It not, would be great to see your way of thinking.
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07 Oct 2009, 19:30
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Well I first started with something like this:
$$P^2/12 = P^2/3*(2^2)$$
As P is a prime number greater than 3, it must be odd, so is $$P^2$$. Now any odd number divided by 3 either gives 1 or 2 as remainder. But we also know that product of any 2 even number is divisible by 4. So I was quite sure that this is going to give me remainder of 1 and not 2. I did some number plugging and it worked so I chose C as the answer.

But then I found something on internet, and then realized that I was going in right direction:
$$P^2 = (P+1)*(P-1) + 1$$. As I stated above, $$(P+1)*(P-1)$$ is surely divisible by 4 as both these numbers are even. Also, one of these 2 numbers is divisible by 3, as these P-1, P and P+1 are consecutive number. Now we can definitely say that $$(P+1)*(P-1)$$ is divisible by 12, and thus $$P^2/12$$ will always give remainder as 1.
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07 Oct 2009, 19:46
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hgp2k wrote:
Well I first started with something like this:
$$P^2/12 = P^2/3*(2^2)$$
As P is a prime number greater than 3, it must be odd, so is $$P^2$$. Now any odd number divided by 3 either gives 1 or 2 as remainder. But we also know that product of any 2 even number is divisible by 4. So I was quite sure that this is going to give me remainder of 1 and not 2. I did some number plugging and it worked so I chose C as the answer.

But then I found something on internet, and then realized that I was going in right direction:
$$P^2 = (P+1)*(P-1) + 1$$. As I stated above, $$(P+1)*(P-1)$$ is surely divisible by 4 as both these numbers are even. Also, one of these 2 numbers is divisible by 3, as these P-1, P and P+1 are consecutive number. Now we can definitely say that $$(P+1)*(P-1)$$ is divisible by 12, and thus $$P^2/12$$ will always give remainder as 1.
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Perfect solution. Here is mine:

(1) not sufficient
(2) not sufficient

(1)+(2) Any prime >3 when divide by 6 can only give remainder 1 or 5 (remainder can not be 2 or 4 because than p would be even, it can not be 3 because p would be divisible by 3)

--> p could be expressed $$p=6n+1$$ or$$p=6n+5$$;

$$p^2=36n^2+12n+1$$ which gives remainder 1 when divided by 12 OR

$$p^2=36n^2+60n+25$$ which also gives remainder 1 when divided by 12

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07 Oct 2009, 19:49
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Good question, +1 buddy!!
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22 Feb 2010, 11:57
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Bunuel wrote:
hgp2k wrote:
Well I first started with something like this:
$$P^2/12 = P^2/3*(2^2)$$
As P is a prime number greater than 3, it must be odd, so is $$P^2$$. Now any odd number divided by 3 either gives 1 or 2 as remainder. But we also know that product of any 2 even number is divisible by 4. So I was quite sure that this is going to give me remainder of 1 and not 2. I did some number plugging and it worked so I chose C as the answer.

But then I found something on internet, and then realized that I was going in right direction:
$$P^2 = (P+1)*(P-1) + 1$$. As I stated above, $$(P+1)*(P-1)$$ is surely divisible by 4 as both these numbers are even. Also, one of these 2 numbers is divisible by 3, as these P-1, P and P+1 are consecutive number. Now we can definitely say that $$(P+1)*(P-1)$$ is divisible by 12, and thus $$P^2/12$$ will always give remainder as 1.
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Perfect solution. Here is mine:

(1) not sufficient
(2) not sufficient

(1)+(2) Any prime >3 when divide by 6 can only give remainder 1 or 5 (remainder can not be 2 or 4 because than p would be even, it can not be 3 because p would be divisible by 3)

--> p could be expressed p=6n+1 or 6n+5;

p^2=36n^2+12n+1 which gives remainder 1 when divided by 12 OR

p^2=36n^2+60n+25 which also gives remainder 1 when divided by 12

Bunuel... could you also elaborat ur approach in deciding... S1 and S2 are insufficient... individually..? Thanks!

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Quote:
Any prime >3 when divide by 6 can only give remainder 1 or 5 (remainder can not be 2 or 4 because than p would be even, it can not be 3 because p would be divisible by 3)

Why do you consider 6... how do you arrive at this?
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22 Feb 2010, 12:12
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jeeteshsingh wrote:
Bunuel wrote:
hgp2k wrote:
Well I first started with something like this:
$$P^2/12 = P^2/3*(2^2)$$
As P is a prime number greater than 3, it must be odd, so is $$P^2$$. Now any odd number divided by 3 either gives 1 or 2 as remainder. But we also know that product of any 2 even number is divisible by 4. So I was quite sure that this is going to give me remainder of 1 and not 2. I did some number plugging and it worked so I chose C as the answer.

But then I found something on internet, and then realized that I was going in right direction:
$$P^2 = (P+1)*(P-1) + 1$$. As I stated above, $$(P+1)*(P-1)$$ is surely divisible by 4 as both these numbers are even. Also, one of these 2 numbers is divisible by 3, as these P-1, P and P+1 are consecutive number. Now we can definitely say that $$(P+1)*(P-1)$$ is divisible by 12, and thus $$P^2/12$$ will always give remainder as 1.
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Perfect solution. Here is mine:

(1) not sufficient
(2) not sufficient

(1)+(2) Any prime >3 when divide by 6 can only give remainder 1 or 5 (remainder can not be 2 or 4 because than p would be even, it can not be 3 because p would be divisible by 3)

--> p could be expressed p=6n+1 or 6n+5;

p^2=36n^2+12n+1 which gives remainder 1 when divided by 12 OR

p^2=36n^2+60n+25 which also gives remainder 1 when divided by 12

Bunuel... could you also elaborat ur approach in deciding... S1 and S2 are insufficient... individually..? Thanks!

Also
Quote:
Any prime >3 when divide by 6 can only give remainder 1 or 5 (remainder can not be 2 or 4 because than p would be even, it can not be 3 because p would be divisible by 3)

Why do you consider 6... how do you arrive at this?

About considering 6: I just explained the common rule that any prime more than 3 can be expressed as $$6n+1$$ or $$6n-1$$ ($$6n+5$$), to plug in $$p^2$$ afterwards.

As for (1) and (2), just plug two different integers: for (1) 4 an 5 (>3) and for (2) 2 and 3 (primes) to see that you'll get two different answers for remainders.

Hope it's clear.
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01 Jun 2011, 07:37
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if just plug in number then whats wrong?
1) p>3, i.e., p= 4,5.....
if p= 4, remainder is 4 and if p=5 remainder 1. thus insufficient

2) p is prime, p = 2, remainder 4, p=5 remainder 1. Insuff.

for C
p= 5, 7 or 11 remainder is 1.
Ans. C
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01 Jun 2011, 11:27
Baten80 wrote:
if just plug in number then whats wrong?
1) p>3, i.e., p= 4,5.....
if p= 4, remainder is 4 and if p=5 remainder 1. thus insufficient

2) p is prime, p = 2, remainder 4, p=5 remainder 1. Insuff.

for C
p= 5, 7 or 11 remainder is 1.
Ans. C

There is no harm at all if you have time crunch. But the case that holds true for 3 numbers can not be surely applied for all.
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14 Jun 2011, 22:16
C with remainder 1 always.

prime = 6n +|- 1 are prime numbers > 3.

prime ^2 = 36n^2 + 12n + 1. thus 1 remainder always.
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17 Oct 2011, 21:15
C.

1 - tons of different remainders, not suff
2 - 2 and 3 squared doesn't even equal 12. after that the remainder is always 1, not suff.
3 - for all prime numbers after 3. when they are squared and divided by 12, the remainder is 1.
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18 Oct 2011, 00:53
'C'

1. P > 3 - Not suff.
2. P is Prime number - Not Suff.

Combined,
- Since P is a prime number greater than 3 then P is surely an odd number.

Tried few prime nums > 3 and got remainder as 1.
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18 Oct 2011, 20:32
C
1. DEFINITELY NOT SUFFICIENT

P> 3, CAN BE ANY NUMBER

2. NOT SUFFICIENT

P IS PRIME NUMBER
2 / 12 REMAINDER IS 2
3 / 12 REMAINDER IS 3

COMBINING BOTH WE HAVE P IS ANY PRIME NUMBER GREATER THAN 3
ANY PRIME NUMBER GREATER THAN 3 IS EXPRESSED IN THE FORM 6K+1 OR 6K-1
SQUARING THEM WE HAVE
36K^2-12K+1 AND 36K^2+12K+1
IN BOTH CASES REMAINDER WHEN DIVIDED BY 12 IS 1
HENCE C
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18 Oct 2011, 21:09
good question Brunel, i arrived at C by plugging in numbers,
but algebraic method sounds good !
thanks
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16 Dec 2011, 06:26
The algebraic method is news to me. Good learning experience. Thanks for the wonderful question Bunuel.
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17 Dec 2011, 23:30
Answer is C. Amit explanation is also nice and straight
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27 Dec 2011, 20:16
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hgp2k wrote:
Well I first started with something like this:
$$P^2/12 = P^2/3*(2^2)$$
As P is a prime number greater than 3, it must be odd, so is $$P^2$$. Now any odd number divided by 3 either gives 1 or 2 as remainder. But we also know that product of any 2 even number is divisible by 4. So I was quite sure that this is going to give me remainder of 1 and not 2. I did some number plugging and it worked so I chose C as the answer.

But then I found something on internet, and then realized that I was going in right direction:
$$P^2 = (P+1)*(P-1) + 1$$. As I stated above, $$(P+1)*(P-1)$$ is surely divisible by 4 as both these numbers are even. Also, one of these 2 numbers is divisible by 3, as these P-1, P and P+1 are consecutive number. Now we can definitely say that $$(P+1)*(P-1)$$ is divisible by 12, and thus $$P^2/12$$ will always give remainder as 1.
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This question also appears on the OG 12th Ed albeit in a PS format.

It's PS Ques 23 on page 155 and goes "If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ?"

Obviously in a PS setting picking any prime and testing for its remainder when divided by 12 is the way to go. OG goes into a theoretical discussion identical to Bunuel’s solution.

However I like the solution based on expressing p^2 as (p+1)*(p-1) + 1 that hgp2k mentions above A LOT MORE !

I usually don't get into the theoetical discussion of this problems with my students, but if I do in the future, I know I'll be using the above.

Thanks a bunch. Oh and since I can't buy you a drink, I gave you a kudos !
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29 Dec 2011, 08:11
+1 C

Not a 700+ question.
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31 Dec 2011, 02:43
Bunuel: i found this one very easy compare to other question that you have discussed in the forum.
Re: Remainder problem   [#permalink] 31 Dec 2011, 02:43

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