It is currently 18 Oct 2017, 22:56

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If p, x, and y are positive integers, y is odd, and p = x^2

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 22 Sep 2005
Posts: 278

Kudos [?]: 241 [17], given: 1

If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

14 Aug 2009, 12:49
17
KUDOS
88
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

40% (02:06) correct 60% (02:17) wrong based on 1935 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
[Reveal] Spoiler: OA

Kudos [?]: 241 [17], given: 1

Manager
Joined: 25 Jul 2009
Posts: 115

Kudos [?]: 262 [42], given: 17

Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
Re: PS: Divisible by 4 [#permalink]

### Show Tags

14 Aug 2009, 13:46
42
KUDOS
27
This post was
BOOKMARKED
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 non-multiple of 4.
=> SUFFICIENT

St2:
x - y = 3
Since y can be any odd number, x could also be either a multiple or a non-multiple of 4.
=> NOT SUFFICIENT

ANS: A
_________________

KUDOS me if I deserve it !!

My GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes

Kudos [?]: 262 [42], given: 17

Math Expert
Joined: 02 Sep 2009
Posts: 41890

Kudos [?]: 128792 [31], given: 12183

Re: PS: Divisible by 4 [#permalink]

### Show Tags

16 Dec 2010, 07:39
31
KUDOS
Expert's post
21
This post was
BOOKMARKED
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 $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ --> $$x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)$$.

So, $$x^2=4(2q+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any case $$2q+1-k^2-k=odd$$ --> $$x^2=4*odd$$ --> 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 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4.

_________________

Kudos [?]: 128792 [31], given: 12183

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17354 [26], given: 232

Location: Pune, India
Re: PS: Divisible by 4 [#permalink]

### Show Tags

19 Dec 2010, 07:49
26
KUDOS
Expert's post
17
This post was
BOOKMARKED
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/divisibility-and-remainders-if-you.html

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.

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [26], given: 232 Intern Joined: 30 May 2013 Posts: 4 Kudos [?]: 8 [8], given: 8 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 15 Sep 2013, 22:13 8 This post received KUDOS 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 Kudos [?]: 8 [8], given: 8 Intern Joined: 02 Sep 2010 Posts: 45 Kudos [?]: 146 [4], given: 17 Location: India Re: PS: Divisible by 4 [#permalink] ### Show Tags 18 Dec 2010, 11:23 4 This post received KUDOS 1 This post was BOOKMARKED 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! Kudos [?]: 146 [4], given: 17 Intern Joined: 14 May 2013 Posts: 12 Kudos [?]: 18 [2], given: 3 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 12 Jun 2013, 10:58 2 This post received KUDOS 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. x-y = 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 Kudos [?]: 18 [2], given: 3 Intern Joined: 25 Jun 2013 Posts: 6 Kudos [?]: 3 [2], given: 0 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 13 Sep 2013, 20:25 2 This post received KUDOS 1 This post was BOOKMARKED 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 Kudos [?]: 3 [2], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [2], given: 232 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 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17354 [2], given: 232

Senior Manager
Joined: 23 Jun 2009
Posts: 360

Kudos [?]: 134 [1], given: 80

Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: PS: Divisible by 4 [#permalink]

### Show Tags

15 Aug 2009, 13:49
1
KUDOS
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

Kudos [?]: 134 [1], given: 80

Director
Joined: 23 Apr 2010
Posts: 573

Kudos [?]: 95 [1], given: 7

Re: PS: Divisible by 4 [#permalink]

### Show Tags

16 Dec 2010, 07:13
1
KUDOS
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.

Kudos [?]: 95 [1], given: 7

Intern
Joined: 25 Mar 2012
Posts: 3

Kudos [?]: 3 [1], given: 1

Re: PS: Divisible by 4 [#permalink]

### Show Tags

16 Jul 2012, 18:19
1
KUDOS
Am i missing something, why cant we take stmt 2 as follows:
squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4

Kudos [?]: 3 [1], given: 1

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17354 [1], given: 232

Location: Pune, India
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

03 May 2017, 05:31
1
KUDOS
Expert's post
VeritasPrepKarishma wrote:
aliasjit wrote:
I am a little confused about solving the problem:

Stmnt 2: x-y =3

we know y is odd.
and if Y is odd as per problem statement y cannot be anything but 1
as x and y both are positive integers.

Therefore x =4 and is divisible by 4.

y is odd, yes, but why must y be 1?

Responding to a pm:
Quote:
because as per statement 2 if x-y =3 and x and y are both +ve integers could y be anything but 1

But as per statement 2, we do not know that x must be 4. x must be even since y is odd and difference between x and y is odd. But will it be 4, we do not know.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [1], given: 232 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [0], given: 232 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 x-y=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 My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17354 [0], given: 232

Manager
Joined: 29 Jun 2011
Posts: 159

Kudos [?]: 24 [0], given: 29

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:):)

Kudos [?]: 24 [0], given: 29

Intern
Joined: 21 Sep 2013
Posts: 9

Kudos [?]: 2 [0], given: 0

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.

Kudos [?]: 2 [0], given: 0

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17354 [0], given: 232

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
Expert's post
1
This post was
BOOKMARKED
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: if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html#p837890
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17354 [0], given: 232

Manager
Joined: 22 Jul 2014
Posts: 130

Kudos [?]: 297 [0], given: 197

Concentration: General Management, Finance
GMAT 1: 670 Q48 V34
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?

Kudos [?]: 297 [0], given: 197

Current Student
Joined: 17 Jul 2013
Posts: 48

Kudos [?]: 33 [0], given: 19

GMAT 1: 710 Q49 V38
GRE 1: 326 Q166 V160
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) non-divisible 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

Kudos [?]: 33 [0], given: 19

Manager
Joined: 22 Jan 2014
Posts: 141

Kudos [?]: 74 [0], given: 145

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) x-y=3 (odd)
even - odd = odd
or odd - even = odd
so insufficient.

Hence, A.
_________________

Illegitimi non carborundum.

Kudos [?]: 74 [0], given: 145

Re: If p, x, and y are positive integers, y is odd, and p = x^2   [#permalink] 31 Aug 2014, 03:38

Go to page    1   2    Next  [ 32 posts ]

Display posts from previous: Sort by