GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 18 Jan 2020, 21:02

### 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

# Given x and y are positive integers such that y is odd, is x divisible

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60480
Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

10 Apr 2015, 04:48
1
9
00:00

Difficulty:

95% (hard)

Question Stats:

51% (02:38) correct 49% (02:46) wrong based on 176 sessions

### HideShow timer Statistics

Given x and y are positive integers such that y is odd, is x divisible by 4?

(1) When (x^2 + y^2) is divided by 8, the remainder is 5.
(2) x – y = 3

Kudos for a correct solution.

_________________
Retired Moderator
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 298
Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

Updated on: 11 Apr 2015, 20:15
4
Bunuel wrote:
Given x and y are positive integers such that y is odd, is x divisible by 4?

(1) When (x^2 + y^2) is divided by 8, the remainder is 5.
(2) x – y = 3

Kudos for a correct solution.

1. (x^2 + y^2) is odd.

y is odd. so x has to be even.

x^2+y^2=8Q+5
x^2+(2a+1)^2=8Q+5
x^2+4a^2+4a+1=8Q+5
x^2=8Q+4-4a^2-4a
x^2=4(2q+1)-4a(a+1)
x^2=4[(2q+1)-a(a+1)]
x^2=4[Odd-Even]
x^2=2^2*Odd^2

Therefore, x CANNOT be divisible by 4 but only by 2.

Sufficient

2. x=y+3

y is odd. So x is even. But we do not know whether x is divisible by 4

Not Sufficient

Originally posted by AmoyV on 10 Apr 2015, 05:18.
Last edited by AmoyV on 11 Apr 2015, 20:15, edited 1 time in total.
Intern
Joined: 10 Jan 2015
Posts: 4
Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

10 Apr 2015, 23:09
1
1) From this statement as the expression leaves a remainder of 5 when divided by 8 and as y is odd so x must be even.
Now if we analyze the different values of x and y it will be as follows:
When
y=3,x=2;y=5,x=2;y=7,x=2 and so on...so as long as y is odd x is equal to 2 to satisfy the given condition and hence sufficient

2)as y is odd so here x is always even and could be 2,4,6,8,.....
So insufficient.

My take is A...
Current Student
Joined: 25 Nov 2014
Posts: 96
Concentration: Entrepreneurship, Technology
GMAT 1: 680 Q47 V38
GPA: 4
Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

Updated on: 11 Apr 2015, 00:02
1
given Y is odd.
1) x^2 + y^2 = 8k+5
8k+5 , for different values of the constant k gives : 5, 13, 21, 29, 37, 45 ...
x^2 + y^2 = 5,13,21,29,37,45 .....
since y = odd, different values of y give :
y=1 : x^2 = 4, 12, 20, 28, 36, 44, ....
y=3 : x^2 = 4, 12, 20, 28, 36, ..... (leave out the negative values)
y=5 : x^2 = 4, 12, 20, ....
Thus x is not divisible by 4. Suff.

2) x-y = 3, and we know y is odd, say 2k+1.
Thus x = 2k+1+3 = 4 +2k.
Now this gives us x = 4,6,8,10.... So x may or may not be divisible by 4.
Not Suff.

Ans A.

Originally posted by SherLocked2018 on 10 Apr 2015, 23:59.
Last edited by SherLocked2018 on 11 Apr 2015, 00:02, edited 1 time in total.
Manager
Joined: 25 Mar 2014
Posts: 125
Location: India
Concentration: Operations, Finance
GMAT Date: 05-10-2015
GPA: 3.51
WE: Programming (Computer Software)
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

11 Apr 2015, 00:00
1
+1 to A.

Putting y = (2n + 1) in y^2, we get (4n^2 + 1 +4n) = 4n(n+1)+1. This expression when divided by 8 always gives reminder 1 for any value of n >=0.

Now, Statement 1 says that When (x^2 + y^2) is divided by 8, the remainder is 5 (4 + 1). Hence x^2 when divided by 8 should give remainder 4, as 1 is coming from y^2. So, x^2 = 8k + 4. x = 2 * [(2k + 1)^(1/2)].
So, its clear that x can never be divided by 4, as [(2k + 1)^(1/2)] can never be even. SUFFICIENT.

B is clearly INUFFICIENT.
Manager
Joined: 25 Mar 2014
Posts: 125
Location: India
Concentration: Operations, Finance
GMAT Date: 05-10-2015
GPA: 3.51
WE: Programming (Computer Software)
Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

11 Apr 2015, 00:06
1
SherLocked2018 wrote:
given Y is odd.
1) x^2 + y^2 = 8k+5
8k+5 , for different values of the constant k gives : 5, 13, 21, 29, 37, 45 ...
x^2 + y^2 = 5,13,21,29,37,45 .....
since y = odd, different values of y give :
y=1 : x^2 = 4, 12, 20, 28, 36, 44, ....
y=3 : x^2 = 4, 12, 20, 28, 36, ..... (leave out the negative values)
y=5 : x^2 = 4, 12, 20, ....
Thus x is not divisible by 4. Suff.

2) x-y = 3, and we know y is odd, say 2k+1.
Thus x = 2k+1+3 = 4 +2k.
Now this gives us x = 4,6,8,10.... So x may or may not be divisible by 4.
Not Suff.

Ans A.

Hi Sherloked,
I think you missed something here. Question is asking about x, not x^2.
x will "never" be divisible by 4.

Thank you.
Manager
Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 106
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

11 Apr 2015, 04:41
1
Bunuel wrote:
Given x and y are positive integers such that y is odd, is x divisible by 4?

(1) When (x^2 + y^2) is divided by 8, the remainder is 5.
(2) x – y = 3

Kudos for a correct solution.

Given: X, Y >0 and Y is ODD
To find : Is X divisible by 4

Solution:

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.

(x^2 + y^2) could take any of the following values 5, 13, 21, 29......

When (x^2 + y^2) = 13 then Y=3 and X=2 and 13 divided by 8 leaves a remainder of 5
X is not Divisible by 4

When (x^2 + y^2) = 29 then Y=5 and X = 2 and 29 divided by 8 leaves a remainder of 5
X is not divisible by 4

Statement 1 is sufficient

Statement 2: x – y = 3

When Y=3 and X = 6 then X-Y = 3
X is not divisible by 4

When Y=1 and X = 4 then X-Y=3
X is divisible by 4

Statement 2 is not sufficient

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15939
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

11 Apr 2015, 16:34
1
Hi AmoyV,

With Fact 1, you have an interesting theory, but you have NO proof that your theory impacts the question that was ASKED. If you do a little more work, you should be able to prove that Fact 1 is actually SUFFICIENT.

Be careful about assuming the "I don't know anything about <blank>" logic means that a Fact is insufficient. DS questions are designed to test you on specific concepts, so it's always better to have proof that you're correct, than to just assume that you are (and possibly miss out on some easy points when your thinking is incomplete).

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Manager
Joined: 15 May 2014
Posts: 61
Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

11 Apr 2015, 18:26
1
Given x and y are positive integers and y is odd

Statement 1:
x^2 + y^2 = 8q + 5
x^2 + (2k+1)^2 = 8q +5
x^2 + 4k^2+4k+1 = 8q + 5
x^2 + 4k^2+4k = 8q + 4
x^2 = 8q + 4 - 4k^2 - 4k
x^2 = 4 [(2q+1) - (k(k+ 1))]
x^2 = 4 [odd - even]
x^2 = 4 * odd square
x = 2 * odd integer; x is not a multiple of 4
Sufficient

Statement 2:
x - y = 3
x = y +3; x could be 4, 6, 8, 10... so x could or couldn't be a multiple of 4
Not Sufficient

Math Expert
Joined: 02 Sep 2009
Posts: 60480
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

13 Apr 2015, 04:12
1
Bunuel wrote:
Given x and y are positive integers such that y is odd, is x divisible by 4?

(1) When (x^2 + y^2) is divided by 8, the remainder is 5.
(2) x – y = 3

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

As of now, we don’t know any specific properties of squares of odd and even integers. However, we do have a good (presumably!) understanding of divisibility. To recap quickly, divisibility is nothing but grouping. To take an example, if we 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 the quotient and 1 is the remainder. For more on these concepts, check out our previous posts on divisibility.

Coming back to our question,
First thing that comes to mind is that if y is odd, y = (2k + 1).

We have no information on x so let’s proceed to the two statements.

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.

The statement tell us something about y^2 so let’s get that.
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), k(k+1) will be even. So 4k(k+1) will be divisible by 4*2 i.e. by 8. So when y^2 is divided by 8, it will leave a remainder 1.

When y^2 is divided by 8, remainder is 1. To get a remainder of 5 when x^2 + y^2 is divided by 8, we should get a remainder of 4 when x^2 is divided by 8. So x must be even. If x were odd, the remainder when x^2 were divided by 8 would have been 1. So we know that x is divisible by 2 but we don’t know whether it is divisible by 4 yet.

x^2 = 8a + 4 (when x^2 is divided by 8, it leaves remainder 4)
x^2 = 4(2a + 1)
So x = 2*?Odd Number

Square root of an odd number will be an odd number so we can see that x is even but not divisible by 4. This statement alone is sufficient to say that x is NOT divisible by 4.

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (Since Even – Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say. This statement alone is not sufficient.

So could you point out the takeaway from this question?

Note that when we were analyzing y, we used no information other than that it is odd. We found out that the square of any odd number when divided by 8 will always yield a remainder of 1.

Now what can you say about the square of an even number? Say you have an even number x.
x = 2a
x^2 = 4a^2

This tells us that x^2 will be divisible by 4 i.e. we can make groups of 4 with nothing leftover. What happens when we try to make groups of 8? We join two groups of 4 each to make groups of 8. If the number of groups of 4 is even, we will have no remainder leftover. If the number of groups of 4 is odd, we will have 1 group leftover i.e. 4 leftover. So when the square of an even number is divided by 8, the remainder is either 0 or 4.

Looking at it in another way, we can say that if a is odd, x^2 will be divisible by 4 and will leave a remainder of 4 when divided by 8. If a is even, x^2 will be divisible by 16 and will leave a remainder of 0 when divided by 8.

Takeaways
– The square of any odd number when divided by 8 will always yield a remainder of 1.
– The square of any even number will be either divisible by 4 but not by 8 or it will be divisible by 16 (obvious from the fact that squares have even powers of prime factors so 2 will have a power of 2 or 4 or 6 etc). In the first case, the remainder when it is divided by 8 will be 4; in the second case the remainder will be 0.
_________________
Current Student
Joined: 25 Nov 2014
Posts: 96
Concentration: Entrepreneurship, Technology
GMAT 1: 680 Q47 V38
GPA: 4
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

13 Apr 2015, 04:31
aniteshgmat1101 wrote:
SherLocked2018 wrote:
given Y is odd.
1) x^2 + y^2 = 8k+5
8k+5 , for different values of the constant k gives : 5, 13, 21, 29, 37, 45 ...
x^2 + y^2 = 5,13,21,29,37,45 .....
since y = odd, different values of y give :
y=1 : x^2 = 4, 12, 20, 28, 36, 44, ....
y=3 : x^2 = 4, 12, 20, 28, 36, ..... (leave out the negative values)
y=5 : x^2 = 4, 12, 20, ....
Thus x is not divisible by 4. Suff.

2) x-y = 3, and we know y is odd, say 2k+1.
Thus x = 2k+1+3 = 4 +2k.
Now this gives us x = 4,6,8,10.... So x may or may not be divisible by 4.
Not Suff.

Ans A.

Hi Sherloked,
I think you missed something here. Question is asking about x, not x^2.
x will "never" be divisible by 4.

Thank you.

Hi anitesh,
Yes, I solved for values of x^2, and then wrote the same, that x is not divisible by 4, since all the values of x^2, give values of x which can not be divisible for 4.

Hope I didnt confuse you
Manager
Joined: 25 Mar 2014
Posts: 125
Location: India
Concentration: Operations, Finance
GMAT Date: 05-10-2015
GPA: 3.51
WE: Programming (Computer Software)
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

13 Apr 2015, 04:38
SherLocked2018 wrote:
aniteshgmat1101 wrote:
SherLocked2018 wrote:
given Y is odd.
1) x^2 + y^2 = 8k+5
8k+5 , for different values of the constant k gives : 5, 13, 21, 29, 37, 45 ...
x^2 + y^2 = 5,13,21,29,37,45 .....
since y = odd, different values of y give :
y=1 : x^2 = 4, 12, 20, 28, 36, 44, ....
y=3 : x^2 = 4, 12, 20, 28, 36, ..... (leave out the negative values)
y=5 : x^2 = 4, 12, 20, ....
Thus x is not divisible by 4. Suff.

2) x-y = 3, and we know y is odd, say 2k+1.
Thus x = 2k+1+3 = 4 +2k.
Now this gives us x = 4,6,8,10.... So x may or may not be divisible by 4.
Not Suff.

Ans A.

Hi Sherloked,
I think you missed something here. Question is asking about x, not x^2.
x will "never" be divisible by 4.

Thank you.

Hi anitesh,
Yes, I solved for values of x^2, and then wrote the same, that x is not divisible by 4, since all the values of x^2, give values of x which can not be divisible for 4.

Hope I didnt confuse you

Sorry my fault..You didnot confuse me at all
VP
Joined: 18 Dec 2017
Posts: 1001
Location: United States (KS)
GMAT 1: 600 Q46 V27
Re: Given x and y are positive integers such that y is odd, is x divisible  [#permalink]

### Show Tags

19 Nov 2019, 09:10
Bunuel wrote:
Given x and y are positive integers such that y is odd, is x divisible by 4?

(1) When (x^2 + y^2) is divided by 8, the remainder is 5.
(2) x – y = 3

Kudos for a correct solution.

Nice Question (So a bump up)

Given: y is positive odd. (1,3,5....)

Asked: x divisible by 4? (x=4/8/12/16.......)

Statement 1:
(1) When (x^2 + y^2) is divided by 8, the remainder is 5.

If y=1 then and (x^2+1) has to leave a remainder of 5 when divided by 8 then x ^2 can be 4 or 12.

Try with y=3 you will find value of x is divisible by 4.

Sufficient.

(2) x – y = 3

Remember y is odd.

y=1 then x=4 Works

y=3 then x=6 . Doesn't Work.

In sufficient.

_________________
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long

Learn from the Legend himself: All GMAT Ninja LIVE YouTube videos by topic
You are missing on great learning if you don't know what this is: Project SC Butler
Re: Given x and y are positive integers such that y is odd, is x divisible   [#permalink] 19 Nov 2019, 09:10
Display posts from previous: Sort by