|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 30 Jul 2009
Posts: 17
Location: Danbury CT
Schools: Wharton, Columbia , Cornell, CMU , Yale
Followers: 2
Kudos [?]:
0
[0], given: 4
|
If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
 20 Aug 2009, 12:29
Question Stats:
0% (00:00) correct
100% (00:00) wrong based on 0 sessions
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
_________________
[b]Make your dream a reality[/b]
|
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
358
[1] , given: 19
|
1
This post received
KUDOS
gulatin2 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) If the remainder is 5 when p is divided by 8, x (or x^2) has to be an even integer since y is already said to be an odd integer. However its not sure whether x could be 2 or 4 or 6 or 8 or so on... If x = 2, y could be any odd integer. Lets say y = 1, p = x^2 + y^2 = 5, which has 5 reminder when it is divided by 8. If x = 2 and y = 3, p = x^2 + y^2 = 13, which has 5 reminder when it is divided by 8. If x = 2, and y = 5, p = x^2 + y^2 = 29, which has 5 reminder when it is divided by 8. But if x = 4, none of the values of y generates 5 reminder when p is divided by 8. For ex: If x = 4, and y = 1, p = x^2 + y^2 = 17, which has 1 reminder when it is divided by 8. If x = 4, and y = 3, p = x^2 + y^2 = 25, which has 1 reminder when it is divided by 8. If x = 4, and y = 5, p = x^2 + y^2 = 41, which has 1 reminder when it is divided by 8. If x = 8, and y = 1, p has 1 reminder when it is divided by 8. If x = 8, and y = 3, p has 1 reminder when it is divided by 8. If x = 8, and y = 5, p has 1 reminder when it is divided by 8. Any value for x i.e. divisible by 4 doesnot produce 5 reminder when p is divided by 8. So it is sufficient to answer that x is not divisible by 4. (2) If x = y+3, p = x^2 + y^2 = (y+3)^2 + y^2. Here y could be 1, or 3 or 5 or 7 or so on.... If y = 1, x = 4...........yes. If y = 3, x = 6...........no. If y = 5, x = 8........... yes.. If y = 7, x = 10.............no.................NSF... So that answer is A.
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Manager
Joined: 10 Aug 2009
Posts: 139
Followers: 3
Kudos [?]:
46
[1] , given: 10
|
1
This post received
KUDOS
yes, statement 2 alone is not sufficient because the only thing we can deduct from it is that x is even but an integer can still be an even but not a multiple of 8.
Statement 1 alone is sufficient:
Assumption: x is a multiple of 4.
If x is a multiple of 4, then x=4m=>x^2=16(m^2)
p=16(m^2)+y^2.
p/8=16(m^2)/8 + y^2/8
16(m^2) is divisible by 8. Therefore the remainder of p when divided by 8 will be equal to the remainer of y^2 when divided by 8.
y is odd => y=2n+1=> y^2=(2n+1)^2=4n^2+4n+1=4n(n+1)+1 We have two subsequent numbers n and n+1. So one of them is even and 4n(n+1) is divisible by 8. y^2=8k+1. When a square of an odd number is divided by 8, the remainder will be 1.
However, we are given that the remainder is 5 => the assumtion that x is a multiple of 4 is wrong. X cannot be a multiple of 4.
Last edited by LenaA on 26 Aug 2009, 11:04, edited 1 time in total.
|
|
|
|
|
|
Intern
Joined: 21 Aug 2009
Posts: 42
Followers: 1
Kudos [?]:
2
[1] , given: 5
|
1
This post received
KUDOS
Given p,x,y>0; y is odd - of the form (2k-1); p = x^2+y^2
Is x divisible by 4?
Stat1: p = 8n+5 => 8n+5 = x^2 + (2k-1)^2 for different values of n and k
rearranging we get x^2 = 4(k-k^2+1+2n)
=> x^2 is a mulitple of 4 and x must be a multiple of 2
Suff. x is not divisible by 4 ( if it is x^2 should be a multiple of 16)
Stat2: (x-y) = 3 or x-(2k-1)= 3
x=2(k+1) => x will be divisible by 4 depending on the value of k
Insuff.
IMO A
|
|
|
|
|
|
Intern
Joined: 16 Mar 2008
Posts: 20
Followers: 0
Kudos [?]:
1
[0], given: 0
|
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
For (1), the minimum value for p should 13(since 8+5/8 = 5). In this case, x can only be 2 (since 2x2 + 3x3 = 13). so x is not divisible by 4. The next possible value for p would be 29(since 8x3+5/8 = 5). Here, x again can only be 3 (since 2x2 + 5x5 = 29). so x is not divisible by 4.
So (1) is sufficient.
For (2), if x-y = 3, it be x should atleast be 4 since y is odd integer(4-1=3). In this case, x is divisible by 4. However, x can be 6(since 6-3=3), in that case its not divisible.
So (2) is not sufficient.
So my answer is A.
|
|
|
|
|
|
Manager
Joined: 12 Aug 2009
Posts: 107
Followers: 3
Kudos [?]:
12
[0], given: 2
|
OK.
(1) - Everyone agrees on this. Suff to prove not divisible by 4
(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.
I will go for 'D'.
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
358
[0], given: 19
|
dolly12 wrote: (2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.
I will go for 'D'. your statemet that "x is even" is not correct. Update: your statemet that "x is even" is correct.
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Manager
Joined: 10 Aug 2009
Posts: 139
Followers: 3
Kudos [?]:
46
[0], given: 10
|
GMAT TIGER wrote: dolly12 wrote: (2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.
I will go for 'D'. your statemet that "x is even" is not correct? I think the statement that x is even is correct. From the statement 2, x-y=3 => x=y+3. 3 is odd and y is odd, then the sum of two odds is even => x is even
|
|
|
|
|
|
Senior Manager
Joined: 27 May 2009
Posts: 286
Followers: 2
Kudos [?]:
23
[0], given: 18
|
LenaA wrote: GMAT TIGER wrote: dolly12 wrote: (2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.
I will go for 'D'. your statemet that "x is even" is not correct? I think the statement that x is even is correct. From the statement 2, x-y=3 => x=y+3. 3 is odd and y is odd, then the sum of two odds is even => x is even agree X can be even ..but it is not sufficient.(2). Kudos to Gmat tiger for a brilliant explanation.
|
|
|
|
|
|
Manager
Joined: 30 Jun 2009
Posts: 52
Followers: 1
Kudos [?]:
3
[0], given: 2
|
All,
I di agree with answer A
However I am having issues understanding the option that x may be 2
To me, p=x^2+y^2
as long as y is odd and positive, its min value is 1
to me x can note be 0 as it is not +ve or -ve
so p>=13
x^2 >= 12
so to me the min value of x is 4.
I am sure that I am missing something here.
Can anyone help?
Thx
|
|
|
|
|
|
Intern
Joined: 24 Aug 2009
Posts: 10
Followers: 0
Kudos [?]:
2
[0], given: 2
|
Stmt I: Suff
Stmt II: InSuff
A
Last edited by deepak115 on 26 Sep 2009, 12:31, edited 1 time in total.
|
|
|
|
|
|
Manager
Joined: 10 Aug 2009
Posts: 139
Followers: 3
Kudos [?]:
46
[0], given: 10
|
defoue wrote: All,
I di agree with answer A
However I am having issues understanding the option that x may be 2
To me, p=x^2+y^2
as long as y is odd and positive, its min value is 1
to me x can note be 0 as it is not +ve or -ve
so p>=13
x^2 >= 12
so to me the min value of x is 4.
I am sure that I am missing something here.
Can anyone help?
Thx No you are not missing. I made a mistake. I will edit my previous post. x cannot be 2 given the fact that x-y=3. The answer is still A though.Thanks for a correction.
|
|
|
|
|
|
Intern
Joined: 27 Aug 2009
Posts: 47
Schools: London Business School
Followers: 3
Kudos [?]:
29
[0], given: 1
|
bml wrote: Given p,x,y>0; y is odd - of the form (2k-1); p = x^2+y^2
Is x divisible by 4?
Stat1: p = 8n+5 => 8n+5 = x^2 + (2k-1)^2 for different values of n and k
rearranging we get x^2 = 4(k-k^2+1+2n)
=> x^2 is a mulitple of 4 and x must be a multiple of 2
Suff. x is not divisible by 4 ( if it is x^2 should be a multiple of 16)
Stat2: (x-y) = 3 or x-(2k-1)= 3
x=2(k+1) => x will be divisible by 4 depending on the value of k
Insuff.
IMO A great explanation
_________________
 Salaries are low in recession. So, working for kudos now.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
x and y are consecutive positive integers and x>y x^2
|
Hermione |
5 |
24 Nov 2006, 23:28 |
 |
20
|
 |
If p, x, and y are positive integers, y is odd, and p = x^2
|
netcaesar |
11 |
14 Aug 2009, 12:49 |
|
|
|
|
If X and Y are positive integers , is y odd ? (1)
|
shashankp27 |
1 |
02 Oct 2011, 15:02 |
|
|
2
|
|
If x and y are positive integers is y odd?
|
devinawilliam83 |
5 |
04 Mar 2012, 23:42 |
|
|
8
|
|
For positive integers x and y, x^2 = 350y. Is y divisible
|
jayaddula |
12 |
09 May 2012, 16:25 |
|
|
|
|
|
|
close
