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

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New post 02 Mar 2012, 09:36
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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.


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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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New post 02 Mar 2012, 10:04
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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.

Answer: C.

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.
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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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New post 02 Mar 2012, 10:22
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..
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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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New post 02 Mar 2012, 10:27
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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New post 27 May 2013, 06:07
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]

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New post 06 Jul 2014, 20:13
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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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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New post 06 Dec 2016, 14:37
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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.


my answer is C too...

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.

answer is C.
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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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New post 27 May 2020, 19:19
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.

Answer: C.

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]

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New post 27 May 2020, 23:55
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.

Answer: C.

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]

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New post 28 May 2020, 00:58
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.

Answer: C.

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.
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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

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