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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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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. DS44402.01
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Re: If x, y, and d are integers and d is odd, are both x and y divisible
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02 Mar 2012, 10:04
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) xy 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) \(xy=dp\) > 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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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
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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
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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
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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) xy is divisible by d. Just a numerical pluginin 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 , (xy) = 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
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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
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07 Jul 2014, 00:01
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: tipsandhintsforspecificquanttopicswithexamples172096.htmlHope it helps.
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Re: If x, y, and d are integers and d is odd, are both x and y divisible
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06 Dec 2016, 14:37
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) xy 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. xy divisible by d suppose d=3 x=5, y=2 > 52 = 3, divisible x=6, y=3 > 63 = 3, divisible one option works, another not. 1+2 since both of them are divisible by 3, sum of x+y and xy 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+yx+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
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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) xy 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) \(xy=dp\) > 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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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
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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) xy 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) \(xy=dp\) > 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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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
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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) xy 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) \(xy=dp\) > 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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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
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