Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 22 Sep 2005
Posts: 262

If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
14 Aug 2009, 12:49
Question Stats:
40% (02:46) correct 60% (02:47) wrong based on 1997 sessions
HideShow timer Statistics
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India

Re: PS: Divisible by 4
[#permalink]
Show Tags
19 Dec 2010, 07:49
netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibilityandremaindersifyou.htmlComing to your question, First thing that comes to mind is if y is odd, \(y^2\) is also odd. If \(y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1\) Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1. Stmnt 1: When p is divided by 8, the remainder is 5. When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4. \(x^2 = 8a + 4\) (i.e. we can make 'a' groups of 8 and 4 will be leftover) \(x^2 = 4(2a+1)\) This implies \(x = 2*\sqrt{Odd Number}\)because (2a+1) is an odd number. Square root of an odd number will also be odd. Therefore, we can say that x is not divisible by 4. Sufficient. Stmnt 2: x  y = 3 Since y is odd, we can say that x will be even (Even  Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient. Answer (A).
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Manager
Joined: 25 Jul 2009
Posts: 114
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN

Re: PS: Divisible by 4
[#permalink]
Show Tags
14 Aug 2009, 13:46
netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 SOL: St1: Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!This means that y^2 MOD 8 = 1 for all y => p MOD 8 = (x^2 + 1) MOD 8 = 5 => x^2 MOD 8 = 4 Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a nonmultiple of 4. => SUFFICIENTSt2: x  y = 3 Since y can be any odd number, x could also be either a multiple or a nonmultiple of 4. => NOT SUFFICIENTANS: A
_________________
KUDOS me if I deserve it !!
My GMAT Debrief  740 (Q50, V39)  My TestTaking Strategies for GMAT  Sameer's SC Notes




Senior Manager
Joined: 23 Jun 2009
Posts: 352
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago

Re: PS: Divisible by 4
[#permalink]
Show Tags
15 Aug 2009, 13:49
Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8



Director
Joined: 23 Apr 2010
Posts: 547

Re: PS: Divisible by 4
[#permalink]
Show Tags
16 Dec 2010, 07:13
Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you.



Math Expert
Joined: 02 Sep 2009
Posts: 50009

Re: PS: Divisible by 4
[#permalink]
Show Tags
16 Dec 2010, 07:39
nonameee wrote: Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you. If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5 > \(p=8q+5=x^2+y^2\) > as given that \(y=odd=2k+1\) > \(8q+5=x^2+(2k+1)^2\) > \(x^2=8q+44k^24k=4(2q+1k^2k)\). So, \(x^2=4(2q+1k^2k)\). Now, if \(k=odd\) then \(2q+1k^2k=even+oddoddodd=odd\) and if \(k=even\) then \(2q+1k^2k=even+oddeveneven=odd\), so in any case \(2q+1k^2k=odd\) > \(x^2=4*odd\) > in order \(x\) to be multiple of 4 \(x^2\) must be multiple of 16 but as we see it's not, so \(x\) is not multiple of 4. Sufficient. (2) x – y = 3 > \(xodd=3\) > \(x=even\) but not sufficient to say whether it's multiple of 4. Answer: A.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 02 Sep 2010
Posts: 35
Location: India

Re: PS: Divisible by 4
[#permalink]
Show Tags
18 Dec 2010, 11:23
maliyeci wrote: Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8 Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question. Even I didn't know about this property of 8. I approached the question in following way: Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem; rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8] if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8. now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1. for y=1; rem(1/8)=1 for y=3; rem(9/8)=1 for y=5;rem(25/8)=1 so you observe this pattern here. coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8; Sufficient Stmt2: y being odd can be accept both 3 and 5 as values and we get different results; thus Insufficient Thus OA is A
_________________
The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!



Intern
Joined: 25 Mar 2012
Posts: 3

Re: PS: Divisible by 4
[#permalink]
Show Tags
16 Jul 2012, 18:19
Am i missing something, why cant we take stmt 2 as follows: squaring xy=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India

Re: PS: Divisible by 4
[#permalink]
Show Tags
16 Jul 2012, 23:24
Eshaninan wrote: Am i missing something, why cant we take stmt 2 as follows: squaring xy=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 The question is: "Is x divisible by 4?" not "Is p divisible by 4?" x is even since y is odd. We don't know whether x is divisible by only 2 or 4 as well.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 14 May 2013
Posts: 10

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
12 Jun 2013, 10:58
netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 1. As p = 8I + 5 we have values of P = 5,13,21,29..... etc .. as y is odd when we solve this p(odd) = x^2 + y^2(odd) x^2 = odd odd = even which can be 2,4,6 ... etc but if we check for any value of p we don't get any multiple of 4. so it say's clearly that x is not divisible by 4. 2. xy = 3 x = y(odd)+3 x is even which can be 2,4,6.. so it's not sufficient .. Ans : A
_________________
Chauahan Gaurav Keep Smiling



Manager
Joined: 29 Jun 2011
Posts: 118
WE 1: Information Technology(Retail)

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
03 Sep 2013, 04:00
Excellent explanation Bunuel & Karishma:):)



Intern
Joined: 25 Jun 2013
Posts: 6

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
13 Sep 2013, 20:25
from first statement p = 8j + 5 Put j as 1, 2,3,4,5... p would be 13, 21,29, 37,45... Now in the formula p= x^2+y^2 put 1,3,5,7 as value of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble by 4. Posted from GMAT ToolKit



Intern
Joined: 30 May 2013
Posts: 4

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
15 Sep 2013, 22:13
I did this question this way. I found it simple.
1. p=x^2+y^2 y is odd p div 8 gives remainder 5. A number which gives remainder 5 when divided by 8 is odd.
so (x^2 + y^2)/8 = oddnumber (x^2 + y^2) = 8 * oddnumber (this is an even number without doubt)
x^2 + y^2 is even. Since y is odd to get x^2+y^2 even x must also be odd.
X is an odd number not divisible by 4
Option A: 1 alone is sufficient



Intern
Joined: 21 Sep 2013
Posts: 8

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
23 Dec 2013, 23:45
For Statement 1: since p when divided by 8 leaves remainder 5.We obtain the following equation p= 8q+5 We know y is odd. If we write p =x^2+y^2 then we get the eqn: x^2+y^2=8q+5 Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd. Then x^2= odd  y^2 i.e x^2=even ie x= even But it's not sufficient to answer the question whether x is a multiple of 4? By this logic i get E as my answer. Statement 2: is insufficient.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
30 Dec 2013, 23:39
Abheek wrote: For Statement 1: since p when divided by 8 leaves remainder 5.We obtain the following equation p= 8q+5 We know y is odd. If we write p =x^2+y^2 then we get the eqn: x^2+y^2=8q+5 Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd. Then x^2= odd  y^2 i.e x^2=even ie x= even But it's not sufficient to answer the question whether x is a multiple of 4?
Your analysis till now is fine but it is incomplete. We do get that x is even but we also get that x is a multiple of 2 but not 4 as explained in the post above: ifpxandyarepositiveintegersyisoddandpx82399.html#p837890
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Current Student
Joined: 22 Jul 2014
Posts: 123
Concentration: General Management, Finance
WE: Engineering (Energy and Utilities)

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
16 Aug 2014, 00:16
For statement 1 , wouldn't plugging in values be a better option?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
18 Aug 2014, 02:43
alphonsa wrote: For statement 1 , wouldn't plugging in values be a better option? No. When you need to establish something, plugging in values is not fool proof. Anyway, in this question, how will you plug in values? You cannot assume a value for x since that is what you need to find. You will assume a value for y and a value for p such that they satisfy all conditions. This itself will be quite tricky. Then when you do get a value for x, you will find that it will be even but not divisible by 4. How can you be sure that this will hold for every value of y and p? When a statement is not sufficient, plugging in values can work  you find two opposite cases  one which answers in yes and the other which answers in no. Then you know that the statement alone is not sufficient. But when the statement is sufficient, it is very hard to prove that it will hold for all possible values using number plugging alone. You need to use logic in that case.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Current Student
Joined: 17 Jul 2013
Posts: 47
GPA: 3.74

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
29 Aug 2014, 05:25
Hi Karishma, Thanks for the explanation to the question. I was just wondering how the answer would change if we change the question stem a little bit. What if the question asks if p (instead of x) is divisible by 4? In this scenario, statement 1 would be sufficient since if something leaves a remainder of 5, it would leave a remainder of 1 upon division by 4 For statement 2, we know that x = y+3, so x is even. If we square it, it would surely be divisible by 4. Now if a number (y^2, which is odd) nondivisible by 4 is added to a number divisible by 4, the result would surely be not divisible by 4. So statement 2 would also be sufficient. Is this reasoning correct? just for practicing the concept



Manager
Joined: 22 Jan 2014
Posts: 176
WE: Project Management (Computer Hardware)

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
31 Aug 2014, 03:38
netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 1) p = 8k+5 (k is a whole number) also p = (x^2+y^2) => (x^2+y^2) mod 8 = 5 any square (n^2) mod 8 follows the following pattern > 1,4,1,0 and then repeats. for getting x^2+y^2 mod 8 = 5 we need to take a 4 and a 1 from the above pattern. at multiples of 4, the remainder is 0. so x can never be divisible by 4. A is sufficient. 2) xy=3 (odd) even  odd = odd or odd  even = odd so insufficient. Hence, A.
_________________
Illegitimi non carborundum.



Director
Joined: 07 Aug 2011
Posts: 540
Concentration: International Business, Technology

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
Show Tags
01 Mar 2015, 09:33
samrus98 wrote: netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 SOL: St1: Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!This means that y^2 MOD 8 = 1 for all y => p MOD 8 = (x^2 + 1) MOD 8 = 5 => x^2 MOD 8 = 4 Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a nonmultiple of 4. => SUFFICIENTSt2: x  y = 3 Since y can be any odd number, x could also be either a multiple or a nonmultiple of 4. => NOT SUFFICIENTANS: Aare there more tricks like this , ODD^2 MOD 8 = 1 ? thanks Lucky
_________________
Thanks, Lucky
_______________________________________________________ Kindly press the to appreciate my post !!




Re: If p, x, and y are positive integers, y is odd, and p = x^2 &nbs
[#permalink]
01 Mar 2015, 09:33



Go to page
1 2
Next
[ 36 posts ]



