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

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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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.
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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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.
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
Do we ignore zero? If say d= 3 and x and y are 3,0 respectively wouldn't the answer be E?
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If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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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.

Check for more here: https://gmatclub.com/forum/tips-and-hint ... 72096.html

Hope it helps.
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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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.

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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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?
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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.
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
Bunuel wrote:

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.

Thank you for the explanation, helps in understanding.
Especially 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 $$k$$):.

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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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Genoa2000
If A->B is true, but !A doesn't implies !B, if d is even.

If d is even, OA will be E.

There are 2 possibilities when d is even.

1) when a or b is divided by d, remainder is d/2.
2) when a or b is divided by d, remainder is 0.

in both cases, a+b and a-b are divisible by d.
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
Genoa2000 wrote:
hemantbafna wrote:
Bunuel wrote:

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.

Thank you for the explanation, helps in understanding.
Especially 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 $$k$$):.

That's true!

What scared me to think in that way was assuming that if A -> B also B -> A. In this situation we have an A <-> B relationship in the rule. Am I right Bunuel VeritasKarishma nick1816?

Can we also state the reverse for this rule?

Yes, you need to be wary of this. Just because A -> B, it doesn't mean B -> A.

Now think in terms of groups of d marbles.
If x is made up of some groups (say 5 grps) of d marbles, and y is made up of some groups (say 7 grps) of d marbles, when you put them together (x + y), you will still get groups (12 grps) of d marbles.
Also, their difference would be a few groups (2 groups) of d marbles too.

Now think of what happens when the sum and difference is in groups of d marbles. Does it mean x and y need to have complete groups of d marbles too?
Say you have total 12 grps of d marbles in x + y and 2 grps of d marbles. The rest of the 10 grps can be equally split into x and y with 5 grps each so x and y will be divisible by d.

But think what happens if you have 12 grps of d marbles in the sum and 3 grps of d marbles as the difference. Then you are left with 9 grps which need to be equally split between x and y. If d is odd you cannot do it but if d is even, you can still split them equally by breaking the grps. Then x and y will not be individually divisible by d.
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If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
If x, y, and d are integers and d is odd, are both x and y divisible by d ?

Rephrased: Is x a multiple of d? Is y a multiple of d?

(1) $$x + y$$ is divisible by d.

let $$d = 3, x = 3, y = 0$$ -- YES

let $$d = 3, x = 2, y = 1$$ -- NO

INSUFFICIENT.

(2) x − y is divisible by d.

let $$d = 3, x = 3, y = 0$$ -- YES

let $$d = 3, x = 4, y = 1$$-- NO

INSUFFICIENT.

(1&2) x + y is divisible by d; x - y is divisible by d:

let $$d = 3, x = 3, y = 0$$ -- YES

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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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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.

Very clear and foolproof approach! Thanks Bunuel
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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.

DS44402.01

Statement 1:
This provides that (x+y)/d = k in which k is some integer. This alone is not sufficient.

Statement 2:
This provides that (x-y)/d = q in which q is some integer. Similar to statement one, this alone is not suffecient.

(1) and (2) together:
Let's start by manipulating the equation in statement 1 to find the value of x in terms of d and k. So x = KD - y. Now let's plug x into the equation created in statement two. So ((kd - y) - y)/d = q. Since q is an integer it must be the case that d divides into (kd-y) and y, which are x and y respectively. Both statements are suffecient.
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If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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.

Are x and y both divisible by d?

If d=1, the answer to the question stem is YES, since every integer is divisible by 1.
The only question is whether the answer to the question stem can be NO.

Statement 1:
Case 1: d=3, x=1 and y=2
In this case, x and y are NOT divisible by d, so the answer to the question stem is NO.
Since the answer can be YES (if d=1) or NO (as in Case 1), INSUFFICIENT.

Statement 2:
Case 2: d=3, x=4 and y=1
In this case, x and y are NOT divisible by d, so the answer to the question stem is NO.
Since the answer can be YES (if d=1) or NO (as in Case 2), INSUFFICIENT.

Statements combined:
Let d=3.
Implication:
x+y is a multiple of 3, with the result that x+y = 3, 6, 9, 12, 15, 18, 21...
x-y is a multiple of 3, with the result that x-y = 3, 6, 9, 12, 15, 18, 21...

Since the sum of x+y and x-y is equal to 2x, the sum of x+y and x-y must be EVEN.

Case 3: x+y=3 and x-y=3, implying that x=3 and y=0
In this case, both x and y are divisible by d=3.
Case 4: x+y=12 and x-y=6, implying that x=9 and y=3
In this case, both x and y are divisible by d=3.
Case 5: x+y=21 and x-y=9, implying that x=15 and y=6
In this case, both x and y are divisible by d=3.

In every case, x and y are both divisible by d.
Implication:
The answer to the question stem is YES.
SUFFICIENT.

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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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Re: If x, y, and d are integers and d is odd, are both x and y divisible [#permalink]
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