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02 Mar 2012, 10:36
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C
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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
(1) x + y is divisible by d.
(2) x − y is divisible by d.
DS44402.01
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02 Mar 2012, 11:04
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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 cannot 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.
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 cannot 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.
27 May 2013, 07:07
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eybrj2
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.
