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

 It is currently 07 Jul 2020, 18:19 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x, y, and d are integers and d is odd, are both x and y divisible

Author Message
TAGS:

### Hide Tags

Manager  Joined: 31 Oct 2011
Posts: 212
If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

4
28 00:00

Difficulty:   55% (hard)

Question Stats: 61% (02:07) correct 39% (01:48) wrong based on 507 sessions

### HideShow timer Statistics

If x, y, and d are integers and d is odd, are both x and y divisible by d ?

(1) x + y is divisible by d.
(2) x − y is divisible by d.

DS44402.01
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

8
1
16
If x, y and d are integers and d is odd, are both x and y divisible by d?

Question asks whether both x and y are multiples of d.

(1) x+y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=2$$, $$y=4$$ and $$d=3$$ then the answer is NO. Not sufficient.

(2) x-y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=4$$, $$y=1$$ and $$d=3$$ then the answer is NO. Not sufficient.

(1)+(2) From (1) $$x+y=dq$$ and from (2) $$x-y=dp$$ --> sum those two equations: $$2x=d(q+p)$$ --> $$x=\frac{d(q+p)}{2}$$. Now, since $$x$$ is an integer and $$d$$ is an odd number then $$\frac{q+p}{2}=integer$$ (d/2 can not be an integer because d is odd) --> $$x=d*integer$$ --> $$x$$ is a multiple of $$d$$. From (1) {multiple of d}+y={multiple of d} --> hence $$y$$ must also be a multiple of $$d$$. Sufficient.

Below might help to understand this concept better.

If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

Hope it's clear.
_________________
##### General Discussion
Manager  B
Joined: 22 Feb 2012
Posts: 80
Schools: HBS '16
GMAT 1: 740 Q49 V42
GMAT 2: 670 Q42 V40
GPA: 3.47
WE: Corporate Finance (Aerospace and Defense)
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

Nice explanation Bunuel - i got it down to C & E and made an educated guess because I couldnt get any example to not work given both conditions..

the part of your explanation that took me a few reads to understand was the

x = D(Q+p)/2 and since d is an odd number (Q+P)/2 must be an integer..
Then i realized since D is odd it couldnt cancel out the 2 below the line to get x to equal an integer..
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

AbeinOhio wrote:
Nice explanation Bunuel - i got it down to C & E and made an educated guess because I couldnt get any example to not work given both conditions..

the part of your explanation that took me a few reads to understand was the

x = D(Q+p)/2 and since d is an odd number (Q+P)/2 must be an integer..
Then i realized since D is odd it couldnt cancel out the 2 below the line to get x to equal an integer..

Making an educated guess for C, based on number plugging is not a bad idea at all, especially if you are running out of time and cannot prove C with 100% certainty.
_________________
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 562
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

eybrj2 wrote:
If x, y and d are integers and d is odd, are both x and y divisible by d?

(1) x+y is divisible by d.

(2) x-y is divisible by d.

Just a numerical plugin-in approach:

From F.S 1, we know that (x+y) is divisible by d. Take x = 5,y = 4,d=3. We get a NO for the question stem.
Again, take x=9,y=3,d=3. We get a YES for the question stem.Insufficient.

From F.S2, just as above, take x=7,y=4,d=3. We get a NO for the question stem.
Again, take x=9,y=6,d=3. We get a YES. Insufficient.

Both taken together, we know that (x+y) = d*m, where m is some integer.
Also , (x-y) = d*n, n is an integer.

Thus, adding we get 2x = d*(m+n). Now, as the LHS is,the RHS has to be even. Also, d is odd. Thus, (m+n) = even --> (m+n) = 2k--> x = d*k and $$\frac{x}{d} = k$$(some integer),now if (x+y) is divisible by d, and x alone is divisible by x, then y has to be divisible by d. Sufficient.

C.
_________________
Manager  B
Joined: 27 May 2014
Posts: 74
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

Do we ignore zero? If say d= 3 and x and y are 3,0 respectively wouldn't the answer be E?
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

1
bankerboy30 wrote:
Do we ignore zero? If say d= 3 and x and y are 3,0 respectively wouldn't the answer be E?

0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself).

Check for more here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html

Hope it helps.
_________________
Board of Directors P
Joined: 17 Jul 2014
Posts: 2422
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

1
eybrj2 wrote:
If x, y and d are integers and d is odd, are both x and y divisible by d?

(1) x+y is divisible by d.

(2) x-y is divisible by d.

1. suppose d=3
x=2, y=1. x+y divisible by 3
x=5, y=1. x+y divisible by 3
one case works, another not.

2. x-y divisible by d
suppose d=3
x=5, y=2 -> 5-2 = 3, divisible
x=6, y=3 -> 6-3 = 3, divisible
one option works, another not.

1+2
since both of them are divisible by 3, sum of x+y and x-y must be divisible by 3
so will the difference between them

x+y+x+y = 2x -> divisible by d. since d is odd, x will 100% be divisible by 3.
x+y-x+y = 2y -> same thing. y is divisible by 3.

Intern  B
Joined: 26 Jun 2019
Posts: 23
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

Bunuel wrote:
If x, y and d are integers and d is odd, are both x and y divisible by d?

Question asks whether both x and y are multiples of d.

(1) x+y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=2$$, $$y=4$$ and $$d=3$$ then the answer is NO. Not sufficient.

(2) x-y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=4$$, $$y=1$$ and $$d=3$$ then the answer is NO. Not sufficient.

(1)+(2) From (1) $$x+y=dq$$ and from (2) $$x-y=dp$$ --> sum those two equations: $$2x=d(q+p)$$ --> $$x=\frac{d(q+p)}{2}$$. Now, since $$x$$ is an integer and $$d$$ is an odd number then $$\frac{q+p}{2}=integer$$ (d/2 can not be an integer because d is odd) --> $$x=d*integer$$ --> $$x$$ is a multiple of $$d$$. From (1) {multiple of d}+y={multiple of d} --> hence $$y$$ must also be a multiple of $$d$$. Sufficient.

Below might help to understand this concept better.

If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

Hope it's clear.

Hello Buenel, With respect to the Concept, Is the reverse of 1 also true? ie., if the sum and difference of x and y is divisible by D, then X and Y are separately divisible by D?
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

babloorudra wrote:
Bunuel wrote:
If x, y and d are integers and d is odd, are both x and y divisible by d?

Question asks whether both x and y are multiples of d.

(1) x+y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=2$$, $$y=4$$ and $$d=3$$ then the answer is NO. Not sufficient.

(2) x-y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=4$$, $$y=1$$ and $$d=3$$ then the answer is NO. Not sufficient.

(1)+(2) From (1) $$x+y=dq$$ and from (2) $$x-y=dp$$ --> sum those two equations: $$2x=d(q+p)$$ --> $$x=\frac{d(q+p)}{2}$$. Now, since $$x$$ is an integer and $$d$$ is an odd number then $$\frac{q+p}{2}=integer$$ (d/2 can not be an integer because d is odd) --> $$x=d*integer$$ --> $$x$$ is a multiple of $$d$$. From (1) {multiple of d}+y={multiple of d} --> hence $$y$$ must also be a multiple of $$d$$. Sufficient.

Below might help to understand this concept better.

If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

Hope it's clear.

Hello Buenel, With respect to the Concept, Is the reverse of 1 also true? ie., if the sum and difference of x and y is divisible by D, then X and Y are separately divisible by D?

No. For example, 5 + 1 = 6 is divisible by 3 but neither 5 nor 1 is.
_________________
Intern  B
Joined: 26 Jun 2019
Posts: 23
Re: If x, y, and d are integers and d is odd, are both x and y divisible  [#permalink]

### Show Tags

Bunuel wrote:
babloorudra wrote:
Bunuel wrote:
If x, y and d are integers and d is odd, are both x and y divisible by d?

Question asks whether both x and y are multiples of d.

(1) x+y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=2$$, $$y=4$$ and $$d=3$$ then the answer is NO. Not sufficient.

(2) x-y is divisible by d --> if $$x=y=d=1$$ then the answer is YES but if $$x=4$$, $$y=1$$ and $$d=3$$ then the answer is NO. Not sufficient.

(1)+(2) From (1) $$x+y=dq$$ and from (2) $$x-y=dp$$ --> sum those two equations: $$2x=d(q+p)$$ --> $$x=\frac{d(q+p)}{2}$$. Now, since $$x$$ is an integer and $$d$$ is an odd number then $$\frac{q+p}{2}=integer$$ (d/2 can not be an integer because d is odd) --> $$x=d*integer$$ --> $$x$$ is a multiple of $$d$$. From (1) {multiple of d}+y={multiple of d} --> hence $$y$$ must also be a multiple of $$d$$. Sufficient.

Below might help to understand this concept better.

If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

Hope it's clear.

Hello Buenel, With respect to the Concept, Is the reverse of 1 also true? ie., if the sum and difference of x and y is divisible by D, then X and Y are separately divisible by D?

No. For example, 5 + 1 = 6 is divisible by 3 but neither 5 nor 1 is.

Thank You for the clarification. Re: If x, y, and d are integers and d is odd, are both x and y divisible   [#permalink] 28 May 2020, 00:58

# If x, y, and d are integers and d is odd, are both x and y divisible  