GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Sep 2018, 02:22

### 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

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

Author Message
TAGS:

### Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8287
Location: Pune, India
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

01 Mar 2015, 20:45
Lucky2783 wrote:
are there more tricks like this , ODD^2 MOD 8 = 1 ?

thanks
Lucky

There can be innumerable properties of numbers and you can not say which one will be useful and which one not on exam day - hence an exercise in learning all such properties is wasted effort. Instead try to figure out how to arrive at them on your own when required.

For example, here y is odd and you need to check divisibility of y^2 by 8.
y = 2a+1

y^2 = (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1

Now, notice that one of a and (a+1) must be even so 4a(a+1) has to be divisible by 8. This means y^2 will always leave remainder 1 when divided by 8.
Focus on developing these inference skills and they will serve you much better on the exam.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Director
Joined: 10 Mar 2013
Posts: 547
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

25 Oct 2015, 05:49
$$[m]$$[/m]
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

x,y,p are integers >0 ,y is odd
(1) p=8a+5 means p is an odd number: $$x^2$$=odd-$$odd^2$$=even, as we are delaing only with integers, x is equal to 0,4,16,36,64,100 etc and is divisible by 4 (0 is also an even number and divisible by 4) SUFFICIENT

(2) x-y=3 means x=odd+odd=even, x=0 is divisible by 4, x=6 is not divisible by 4, two different answers,so NOT SUFFICIENT
btw. the difference between (1) and (2) when $$x^2$$=even must be divisible by 4, but when x=even it could be divisible by 4.
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Intern
Joined: 09 Jul 2015
Posts: 2
Concentration: Entrepreneurship, Marketing
GPA: 3.99
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

22 Nov 2015, 03:43
Can't we think of it in a simpler way, without using much properties.

Statement 1 says that "When p is divided by 8, the remainder is 5", therefore we can infer that p is an odd number.
now since it is given that y is odd therefore square of y will also be odd, by this we come to know that x has to be an even number.
P(ODD)=x(EVEN)+y(ODD)

Now, we take the smallest value of a positive even integer for x, i.e. 2.
2^2=4, therefore YES.
By rule, square of any any even integer is always divisible by 4. Therefore statement 1 is sufficient

Statement 2 states that x – y = 3,
if X=4 and Y=1, then Yes
If X=6 and Y=3, then No

hence insufficient.

Im not a pro so might end up doing something wrong. Please do correct me if i make any mistake here.

Thanks
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6227
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

26 Nov 2015, 07:58
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

The question is actually asking whether when x=4m?(m;positive integer), p=(4m)^2+odd^2=odd?
There are 3 variables (p,x,y) but only 2 equations are given by the conditions, so there is high chance (D) will be the answer
For condition 1, p=8q+5=odd(q: positive integer) p=5,13,21......., When y=1,3,5.... and x=2,6,10.... the answer is always 'no' so the condition is sufficient
For condition 2, the answer becomes 'yes' for x=4,y=1, but 'no' for x=6, y=3, so this is insufficient and the answer becomes (A).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Intern
Joined: 26 Apr 2017
Posts: 6
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

27 Apr 2017, 05:59
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.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8287
Location: Pune, India
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

27 Apr 2017, 06:07
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?
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8287
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
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

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Manager
Joined: 30 Dec 2015
Posts: 89
GPA: 3.92
WE: Engineering (Aerospace and Defense)
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

19 Jun 2017, 19:33
y is odd => y = 2n+1
y^2 = 4*even + 1 or 8n + 1; where n can be any positive integer
Statement 1:
P = 8a + 5
8a + 5 = x^2 + 8n + 1
x^2 = 8(a-n) + 4 = 4(2a – 2n + 1) = 4(odd)
x = 2(odd) ; x = 2, 6, 10, 14……
Hence X is not divisible by 4
SUFFICIENT
Statement 2:
x-y = 3; x = 3 – odd; x = even. X can be 2, 4, 6, 8; NOT SUFFICIENT
_________________

If you analyze enough data, you can predict the future.....its calculating probability, nothing more!

Manager
Joined: 04 May 2014
Posts: 161
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

18 Aug 2017, 22:09
p=x²+y²
Y=odd

question is is x/4=integer
1) p/8 gives a reminder of 5
or p=q8+5(q is the quotient)
P can be =5, 13,18,23,28,34,39,44.......
let us check out options by plugin in above nos
p=5
ie 5=x²+y² now y can only be 1 and x=2. which is not divisible by 4
13=x²+y²=(2)²+(3)² again not divisible by 4
18=(3)²+(3)² again 3 is not divisible by 4
23= (>4<5)²+(3)² again not divisible by 4
Statement is sufficient
2) X-Y=3
X-odd=odd
X has to be even. It can be 2 which is not divisible by 4 or 4 which is divisible by 4
Statement is not sufficient
Intern
Joined: 23 Dec 2014
Posts: 9
If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

20 Aug 2017, 10:40
siddharthmk wrote:
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

Hello All,

A quick observation regarding text highlighted above: I believe we can NOT write like this. Lets say p = x^2+y^2 = 5 or 13 or 21 or 29 so on.

Then we can Not write 5/8 = odd number or 13/8 = odd number Isn't? Rather we can write 13/8 = 1 + 5/8 i.e. Dividend/Divisor = Quotient + Reminder/Divisor . Any comments maths expert Bunel/Karishma ?
Senior Manager
Joined: 09 Feb 2015
Posts: 365
Location: India
Concentration: Social Entrepreneurship, General Management
GMAT 1: 690 Q49 V34
GMAT 2: 720 Q49 V39
GPA: 2.8
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

04 Sep 2017, 05:13
Bunuel wrote:
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.

For statement 1 , can we not directly say that since the rem is 5 when divided by 8 ,p wll be odd.
Since y is already odd => x is odd . hence not divisible by 4?
Senior Manager
Joined: 02 Apr 2014
Posts: 477
GMAT 1: 700 Q50 V34
If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

15 Nov 2017, 12:00
$$y$$ is odd , Let $$y = 2k + 1$$ (any odd number can be expressed like this)

Statement 1:
Now $$p % 8 = 5$$ => $$(x^2 + y^2 => x^2 + (2k + 1)^2)$$ % 8 = 5
=> $$(x^2 + 4k^2 + 4k + 1)$$ % 8 = 5
=> $$(x^2 + 4k^2 + 4k -4 + 4 + 1)$$ % 8 = 5 (since 5 is remainder)
$$(x^2 + 4k^2 + 4k - 4)$$ % 8 = 0
$$(x^2 + 4(k^2 + 4k - 1))$$ % 8 = 0
now if $$k = odd => k^2 + 4k - 1$$ is odd
now if $$k = even => k^2 + 4k - 1$$ is still odd
so, $$(x^2 + 4 * odd)$$ % 8 = 0
if $$x$$ were mulitple of 4, then $$x = 4m$$
=> $$(16m^2 + 4 * odd)$$ % 8=> this won't be zero as $$16m^2$$ is divisible by 8, but $$4*odd$$ is not.
so x cannot be 4m (or multiple of 4)

sufficient

Statement 2:
Clearly not sufficient

Intern
Joined: 08 Jul 2018
Posts: 1
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

21 Jul 2018, 03:08
Bunuel wrote:
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.

Can someone tell me where I went wrong? I did:
p=8q+5=x^2+y^2
then x^2 = 8q + 5 - y^2
8q is even, 5 is odd, y^2 is odd (since y is odd)
so x^2 = even + odd - odd = even
Intern
Joined: 05 Mar 2015
Posts: 49
Location: Azerbaijan
GMAT 1: 530 Q42 V21
GMAT 2: 600 Q42 V31
GMAT 3: 700 Q47 V38
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

06 Aug 2018, 02:03
From the statement one we know that p=8t+5
I listed possible values for p:
5, 13, 21, 29, 37, 45... from the list I found only 45 as a sum of two squares (6^2+3^2) and chose answer A.

But I was not sure 45 is the only number that satisfies p=x^2+y^2. How to know whether a number other than 45 satisfies p=x^2+y^2 and p=8t+5 requirement
Intern
Joined: 01 Jun 2011
Posts: 20
Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

### Show Tags

06 Sep 2018, 02:20
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: http://gmatclub.com/forum/if-p-x-and-y- ... ml#p837890

Is the following solution correct?

x2=8q+5+y2

Since x is an integer the right hand side must also be an integer. Since right hand side is even it has to be a perfect even square to equal the left side.
Y cannot assume just any value it has to assume only those which will make the right side an even perfect square. So x must be even and and a multiple of 4.
Re: If p, x, and y are positive integers, y is odd, and p = x^2 &nbs [#permalink] 06 Sep 2018, 02:20

Go to page   Previous    1   2   [ 36 posts ]

Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.