Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
If p is a positive integer, what is a remainder when p^2 is [#permalink]
07 Oct 2009, 18:37
4
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
45% (medium)
Question Stats:
68% (07:42) correct
32% (01:07) wrong based on 222 sessions
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.
Re: Remainder problem [#permalink]
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.
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.
Re: Remainder problem [#permalink]
07 Oct 2009, 19:30
9
This post received KUDOS
1
This post was BOOKMARKED
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
Re: Remainder problem [#permalink]
07 Oct 2009, 19:46
7
This post received KUDOS
Expert's post
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. ________________________________ 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. _________________
Re: Remainder problem [#permalink]
22 Feb 2010, 11:57
1
This post was BOOKMARKED
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. ________________________________ 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|
Re: Remainder problem [#permalink]
22 Feb 2010, 12:12
Expert's post
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. ________________________________ 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.
Re: Remainder problem [#permalink]
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.
Re: Remainder problem [#permalink]
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. _________________
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
Re: Remainder problem [#permalink]
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
Re: Remainder problem [#permalink]
27 Dec 2011, 20:16
1
This post received KUDOS
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. ________________________________ 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 ! _________________
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...