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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.

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.

Answer with explanation will follow.

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.

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. ________________________________ Consider KUDOS for good posts

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. ________________________________ Consider KUDOS for good posts

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

So answer C. Thanks for your reply +1. _________________

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. ________________________________ Consider KUDOS for good posts

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

So answer C. Thanks for your reply +1.

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? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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. ________________________________ Consider KUDOS for good posts

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

So answer C. Thanks for your reply +1.

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.

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.

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. _________________

GMAT is an addiction and I am darn addicted

Preparation for final battel: GMAT PREP-1 750 Q50 V41 - Oct 16 2011 GMAT PREP-2 710 Q50 V36 - Oct 22 2011 ==> Scored 50 in Quant second time in a row MGMAT---- -1 560 Q28 V39 - Oct 29 2011 ==> Left Quant half done and continued with Verbal. Happy to see Q39

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

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. ________________________________ Consider KUDOS for good posts

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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