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If x, y and z are integers and xy + z is an odd integer, is x an even
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If x, y and z are integers and xy + z is an odd integer, is x an even integer? (1) xy + xz is an even integer (2) y + xz is an odd integer
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Originally posted by Aleehsgonji on 28 Jul 2009, 22:25.
Last edited by Bunuel on 17 Jun 2019, 08:35, edited 2 times in total.
Edited the question and added the OA.




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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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20 Mar 2014, 06:09
Mountain14 wrote: jlgdr wrote: Aleehsgonji wrote: If x, y and z are integers and xy + z is an odd integer, is x an even integer?
(1) xy + xz is an even integer (2) y + xz is an odd integer Odd/Even questions can be usually solved quite easily if one tries some operations with the statements We want to know if x is even integer We are given that xy+z is odd Statement 1 xq + xz is even Subtracting z(x+1) is odd
Therefore, x+1 should be odd and x should be evenSufficient Statement 2 Not sufficient Answer is A Just my 2c Cheers J I am not clear with the red part. When you subtract \(xy + z=odd\) from \(xy+xz=even\) you'll get: \(xzz=evenodd=odd\) > \(z(x1)=odd\). For the product of two integers to be odd, both of them must be odd > \(z\) and \(x1\) are odd. If \(x1=odd\), then x must be even: \(x1=xodd=odd\) > \(x=odd+odd=even\). Hope it's clear.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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28 Jul 2009, 23:14
IMO A 1. xy + xz is an even integer  SUFFICIENT Given: xy + z is odd ...(i) xy + xz is even ...(ii) subtracting (ii) from (i) we get xz  z, which should be odd (* since odd  even = odd) => z(x1) is odd => both z and (x1) is odd => since (x1) is odd, x must be even. 2. y + xz is an odd integer INSUFFICIENT Given: xy + z is odd ...(i) y + xz is odd ...(ii) subtracting (ii) from (i) we get xy + z  y  xz = (x1)(yz) , which should be even => either (x1) is even or (yz) is even ....insufficient to determine
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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26 Sep 2009, 22:52
You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear. Or you can proceed algebraically  notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz  (xy + z) = xz  z = z(x1) is odd. Since this is a product, z must be odd, and x1 must be odd, so x is even. Sufficient. For Statement 2, all the letters could be odd, so not sufficient.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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27 Sep 2009, 09:58
IanStewart wrote: You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.
Or you can proceed algebraically  notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz  (xy + z) = xz  z = z(x1) is odd. Since this is a product, z must be odd, and x1 must be odd, so x is even. Sufficient.
For Statement 2, all the letters could be odd, so not sufficient. Fr St2, y+xz odd xy+z odd => y+zx+xy+z even => y(x+1)+z(x+1) even => (y+z)(x+1) event x+1 can be odd or even means that x can be even or odd, insuff



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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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29 Sep 2009, 19:26
Took me less than a min
if XY+Z is odd these are the possiblities
X Y Z
E E O O E O E O O O O E
so substituting these values in Option A gives the solution write away as Even. in B we have 2 diff results so
Answer is A



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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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31 Jan 2014, 09:31
Aleehsgonji wrote: If x, y and z are integers and xy + z is an odd integer, is x an even integer?
(1) xy + xz is an even integer (2) y + xz is an odd integer Odd/Even questions can be usually solved quite easily if one tries some operations with the statements We want to know if x is even integer We are given that xy+z is odd Statement 1 xq + xz is even Subtracting z(x+1) is odd Therefore, x+1 should be odd and x should be even Sufficient Statement 2 Not sufficient Answer is A Just my 2c Cheers J



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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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21 Mar 2014, 20:46
Odd(O) Even (E) given: x,y,z integers xy+z=O so only the following scenarios can fulfill the constraints a) EO+O b) EE+O c) OE+O d) OO+E question: x=E? 1) x(y+z)=E i. (E)(O+O) > fits scenario a >yes, x can be even ii. (O)(E+E) > n/a  doesn't fit any scenarios iii. (O)(O+O) > n/a  doesn't fit any scenarios stop testing, x can't be odd, sufficient 2) y+xz = O i. E+(O)(O) > fits scenario a >yes, x can be even ii. O+(E)(E) > n/a  doesn't fit any scenarios iii. O+(O)(E) > fits scenario d >no, x can be odd stop testing, x can be either even or odd insufficient A
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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27 May 2016, 20:38
rahsin wrote: If x, y, z are integers and xy+z is an odd integer, is x an even integer.
1. xy +xz is an even integer
2. y+ xz is an odd integer Statement 1xy+xz is even xy+z is odd Subtract the two. xy+xzxyz = xzz xzz will be odd (even  odd will always result in odd) z(x1) will be odd This is only possible when both z and (x1) are odd. x1 is odd. This means that x is even. Sufficient. Statement 2y+xz is odd xy+z is odd Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z). (x+1)(y+z) will be even (odd+odd is even) Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result) or x+1 can be odd and y+z can be even (Even * odd will result in an even result) or both (x+1) and (y+z) can be even (Even * even will result in an even result) So x+1 can be both odd and even. In other words, x can be both even and odd. This is only possible when both z and (x1) are odd. x1 is odd. This means that x is even. NOT Sufficient. A is the answer.



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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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31 Jan 2017, 19:03
(1) Since xy + z is odd and xy +xz is even, subtracting we can deduce that zxz = z(1x) is odd. Thus 1  x is odd, so x must be even. SUFF (2) SUbtracting, we see that xy  y + z  xz = y(x  1)  z(x  1) =(y  z)(x  1) is even. If y  z is odd, x  1 is even and thus x is odd. On the other hand, if y  z is even, x is even. NOT SUFF Hence A.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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27 Dec 2017, 21:55
Aleehsgonji wrote: If x, y and z are integers and xy + z is an odd integer, is x an even integer?
(1) xy + xz is an even integer (2) y + xz is an odd integer Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Since we have 3 variables (x and y) and 1 equations, C is most likely to be the answer and so we should consider 1) & 2) first. Conditions 1) & 2) xy + xz : even ⇒ xy:even, xz=even  xy:odd, xz:odd y + xz : odd ⇒ y: odd, xz=even  y: even, xz: odd xy + z: odd ⇒ xy: even, z:odd  xy:odd, z:even ⇒ x:even, y:odd, z:odd Both conditions together are sufficient. Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A). Condition 1) xy + xz : even ⇒ xy:even, xz=even  xy:odd, xz:odd xy + z: odd ⇒ xy: even, z:odd  xy:odd, z:even ⇒ x:even, y:?, z:odd The condition 1) is sufficient since we can determine x is even. Condition 2) y + xz : odd ⇒ y: odd, xz=even  y: even, xz: odd xy + z: odd ⇒ xy:odd, z:even  xy: even, z:odd x = 2, y = 1, z = 1  x = 1, y = 2, z = 1 The condition 2) is not sufficient. Therefore, A is the answer. Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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12 Jul 2018, 20:43
Aleehsgonji wrote: If x, y and z are integers and xy + z is an odd integer, is x an even integer?
(1) xy + xz is an even integer (2) y + xz is an odd integer Please find solution as attached. Answer: Option A
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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13 Jul 2018, 04:28
Aleehsgonji wrote: If x, y and z are integers and xy + z is an odd integer, is x an even integer?
(1) xy + xz is an even integer (2) y + xz is an odd integer Statement 1: Test whether it's possible that x is odd. Let x=1. Substituting x=1 into xy+z = ODD, we get: y+z = ODDSubstituting x=1 into xy+xz = EVEN, we get: y+z = EVENThe equations in red contradict each other, implying that it is not possible that x is odd. Thus, x must be even. SUFFICIENT. Statement 2: Test whether it's possible that x is odd. Let x=1. Substituting x=1 into xy+z = ODD, we get: y+z = ODDSubstituting x=1 into y+xz = ODD, we get: y+z = ODDSince the equations in green are the same, it's possible that x is odd. Test whether it's possible that x is even. Let x=2. Substituting x=2 into xy+z = ODD, we get: 2y+z = ODDSubstituting x=2 into y+xz = ODD, we get: y+2z = ODDThe equations in green are both viable if y=1 and z=1, implying that it's possible that x is even. Since the answer to the question stem is NO in the first case but YES in the second case, INSUFFICIENT.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even
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03 Jul 2019, 16:40
Hi All, We're told that X, Y and Z are INTEGERS and (X)(Y) + Z is an ODD integer. We're asked if X is an EVEN integer. This is a YES/NO question and can be solved by either TESTing VALUES or using Number Properties. While it certainly appears more complex than a typical DS prompt, the basic Number Property rules involved are just about multiplication and addition, so you might find it easiest to think in terms of those patterns (although this question is stepheavy and you do have to be thorough with your work and not miss any of the options). To start, since (X)(Y) + Z is an ODD, we can 'map out' the limited number of possibilities for the 3 variables. We could have... (X)(Y) = Even, Z = Odd (X)(Y) = Odd, Z = Even From here, we then need to subdivide (X)(Y): IF (X)(Y) is Even, then either both are Even OR one is Odd and the other is Even (this includes the option in which one is 0 and the other could be any integer). IF (X)(Y) is Odd, then both MUST be Odd This ultimately means that these are the only 4 options for the three variables: X = Even, Y = Even, Z = Odd X = Even, Y = Odd, Z = Odd X = Odd, Y = Even, Z = Odd X = Odd, Y = Odd, Z = Even Having this list should make working through the two Facts a lot easier. (1) (X)(Y) + (X)(Z) is an EVEN integer Using the 4 options above, (X)(Y) + (X)(Z) would be.... (E)(E) + (E)(O) = Even... This IS a match (E)(O) + (E)(O) = Even... This IS a match (O)(E) + (O)(O) = Odd... This is NOT a match (O)(O) + (O)(E) = Odd... This is NOT a match Two of the options fit Fact 1  and they BOTH require that X be EVEN, so the answer to the question is ALWAYS YES. Fact 1 is SUFFICIENT (2) Y + (X)(Z) is an ODD integer Using the 4 options above, (Y) + (X)(Z) would be.... (E) + (E)(O) = Even... This is NOT a match (O) + (E)(O) = Odd... This IS a match (E) + (O)(O) = Odd... This IS a match (O) + (O)(E) = Odd... This IS a match In the three options that fit Fact 2, X is sometimes Even (a "Yes" answer) but sometimes Odd (a "No" Answer). Fact 2 is INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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If x, y and z are integers and xy + z is an odd integer, is x an even
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07 Jul 2019, 14:03
adiagr wrote: rahsin wrote: If x, y, z are integers and xy+z is an odd integer, is x an even integer.
1. xy +xz is an even integer
2. y+ xz is an odd integer Statement 1xy+xz is even xy+z is odd Subtract the two. xy+xzxyz = xzz xzz will be odd (even  odd will always result in odd) z(x1) will be odd This is only possible when both z and (x1) are odd. x1 is odd. This means that x is even. Sufficient. Statement 2y+xz is odd xy+z is odd Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z). (x+1)(y+z) will be even (odd+odd is even) Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result) or x+1 can be odd and y+z can be even (Even * odd will result in an even result) or both (x+1) and (y+z) can be even (Even * even will result in an even result) So x+1 can be both odd and even. In other words, x can be both even and odd. This is only possible when both z and (x1) are odd. x1 is odd. This means that x is even. NOT Sufficient. A is the answer.Both addition or subtraction of equations can work. Order to equations (Eq 2  Eq 1 or Eq 1  Eq 2) for subtraction doesn't matter. For statement 1: Instead of subtracting, if we add then: xy+z = odd xy +xz = even xy + z + xy + xz = odd + even => 2xy + z(1+x) = odd => even + z(1+x) = odd => z(1+x) = odd => 1+x = odd => odd + x = odd => x = even => Sufficient For statement 1: Instead of adding, if we subtract then: xy+z = odd y+ xz = odd xy + z  y  xz = odd  odd => y(x1) + z(1x) = even => y(x1) z(x1) = even => (x1)(yz) = even => (x1) = even or odd => If (x1) = even, then x = odd. If (x1) = odd, then x = even. => Not Sufficient
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If x, y and z are integers and xy + z is an odd integer, is x an even
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