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The positive integers x, y, and z are such that x is a [#permalink]
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The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? (1) xz is even (2) y is even.
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Last edited by Bunuel on 25 Feb 2012, 02:35, edited 1 time in total.
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Re: positive integers [#permalink]
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amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even
2) y is even.
Please explain. thanks. Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought. When we say, "The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.", it implies that if x or y is even, z will be even. e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even. Once this makes sense to you, it will take 10 secs to arrive at the solution. Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even. Stmnt 2: y is even, so z must be even
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Re: positive integers [#permalink]
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amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even. Given: x is a factor of y > \(y=mx\), for some nonzero integer \(m\); y is a factor of z > \(z=ny\), for some nonzero integer \(n\); So, \(z=mnx\). Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even (1) \(xz\) even > either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient. (2) \(y\) even > as \(z=ny\) then as one of the multiples of z even > z even. Sufficient. Answer: D.
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Re: positive integers [#permalink]
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amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even 2) y is even.
Please explain. thanks. y/x = k where k is an integer. y = xk ....................i z/y = m where m is an integer. z = ym = xkm .....................ii If a factor is even, then the source of the factor must be even. 1) If xz is even, z must be even because x may or may not be an even because x is a factor of z but z must be even. SUFF. 2) If y is even, z must be even because y is a factor of z. SUFF.. D..
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Re: positive integers [#permalink]
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VeritasPrepKarishma wrote: amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even
2) y is even.
Please explain. thanks. Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought. When we say, "The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.", it implies that if x or y is even, z will be even. e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even. Once this makes sense to you, it will take 10 secs to arrive at the solution. Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even. Stmnt 2: y is even, so z must be even awesome. your explanations and bunuel as well, are amazing thanks
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Re: The positive integers x, y, and z are such that x is a [#permalink]
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27 Feb 2012, 05:20
dchow23 wrote: if we plug and play, why can't we test X = 1? then y/n.. Plug in method would be far more painful for this question (and most other questions in my opinion). Think how you would go about it: Checking whether stmnt 1 is enough: xz is even If x = 1, y = 1 and z = 2 (so that xz is even), then z is even. If x = 2, y = 2 and z = 4, z is again even. Then you start thinking if you can take some values such that xz is even but z is not... Now you start using logic... Wouldn't you say it is far better to use logic in the first place itself?
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Re: positive integers [#permalink]
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fozzzy wrote: Bunuel wrote: (1) \(xz\) even > either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
Can you please provide a numerical example for this part it isn't very clear. Thanks in advance! I thought it was insufficient as there were multiple cases either X or Y even or both.... We know that x is a factor of \(y \to y = Ix\) Again, \(z = yI^'\). Now, given that xz  even. Case I:Assume that x = even , z = odd. Now, as x is even, y = even(I can be odd/even,doesn't matter). Again, as y is even, z HAS to be even(\(I^'\) is odd/even, doesn't matter). Thus, if x is even, z IS even. Numerical Example :y = 2*I(x=2). \(z = 6*I^'\). z IS even. Case II : z is even OR (x and z) both are even. Hope this helps.
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Re: positive integers [#permalink]
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12 Dec 2010, 10:22
Bunuel, you always try to solve the questions algebraically, don't you?
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Re: positive integers [#permalink]
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12 Dec 2010, 19:36
I like to thing of the boxes method. If you draw them out, then x is inside y which is inside z. zx, a 2 will exist inside the box of either z or x (which is itself inside z) so YES y a 2 will exist inside the box of y which is itself z so YES



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Re: positive integers [#permalink]
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17 Mar 2011, 23:07
Easy if you realize the following: When a is a factor of b AND b is a factor of c THEN a is a factor of c as well.
Hence when either one of these numbers is even, the other has to be even too.. D



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Re: positive integers [#permalink]
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17 Mar 2011, 23:52
z = ky y = mx so z = (km)xy (1) > xz is eve means at least x or z is even, and if x = even, then z is also even as it has an even factor. 2 > y is even so z having an even factor is even too. Answer D
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Re: positive integers [#permalink]
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15 Apr 2011, 05:24
At first, mistook "factor" for "multiple" came with answer E. Later, understood that the problem was so easy...just plug and play !!



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Re: The positive integers x, y, and z are such that x is a [#permalink]
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24 Feb 2012, 10:06
if we plug and play, why can't we test X = 1? then y/n..



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Re: The positive integers x, y, and z are such that x is a [#permalink]
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24 Feb 2012, 10:46
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? 1) xz is even 2) y is even. Please explain. thanks.[/quote]
Ans. let us take any 3 numbers, say x=3,y=18,z=54, or x=2,y=4,z=20 1)if xz is even then it means that either x or z is even,say that x is even, now there is no even number which is a factor of odd number, so z is definitely even, now if x is odd as in the above case, still then we can point out that z is even. 2)if y is even then it is clear that z will be even. Thus this question could be answered by any of the two questions.



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Re: The positive integers x, y, and z are such that x is a [#permalink]
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22 Jan 2013, 22:49
amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even (2) y is even.
What is given? \(z = y*N\) \(y = x*R\) 1. \(xz = 2*I\) If x is even, then z is even since \(z = x*R\) If z is even, then z is even. SUFFICIENT! 2. \(y = 2*I\) ==> \(z = 2*I*N\) Definitely EVEN SUFFICIENT! Answer: D
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Re: positive integers [#permalink]
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03 Aug 2013, 02:04
Bunuel wrote: (1) \(xz\) even > either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
Can you please provide a numerical example for this part it isn't very clear. Thanks in advance! I thought it was insufficient as there were multiple cases either X or Y even or both....
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Re: positive integers [#permalink]
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10 Oct 2013, 15:01
Bunuel wrote: amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even. Given: x is a factor of y > \(y=mx\), for some nonzero integer \(m\); y is a factor of z > \(z=ny\), for some nonzero integer \(n\); So, \(z=mnx\). Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even (1) \(xz\) even > either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient. (2) \(y\) even > as \(z=ny\) then as one of the multiples of z even > z even. Sufficient. Answer: D. Perfect explanation. Remember the factor foundation rule. Also, other properties of factors that might be helpful to have in mind. Just to remind you, The factor foundation rule states that "if a is a factor of b, and b is a factor of c, then a is a factor of c" Also, if 'a' is a factor of 'b', and 'a' is a factor of 'c', then 'a' is a factor of (b+c). In fact, 'a' is a factor of (mb + nc) for all integers 'm' and 'n' If 'a' is a factor of 'b' and 'b' is a factor of 'a', then 'a=b' If 'a' is a factor of 'bc' and gcd (a,b) = 1, then 'a' is a factor of 'c' If 'p' is a prime number and 'p' is a factor of 'ab' then 'p' is a factor of 'a' or 'p' is a factor of 'b' In other words,, any integer is divisible by all of its factors and it is also divisible by all of the factors of its factors Show your appreciation in Kudos. Hope it helps



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