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
Could someone please explain how to get to the answer as quickly as possible?
Thanks
tricky odd/even integer question
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 418
- Joined: Wed Jun 11, 2008 5:29 am
- Thanked: 65 times
There are 4 cases that will satisfy xy + z = odd
(i) x = even, y = even, z = odd
(ii) x = even, y = odd, z = odd
(iii) x = odd, y = even, z = odd
(iv) x = odd, y = odd, z = even
is x even?
Statement 1:
xy + xz = even
x(y + z) = even
case (a): x = odd, (y + z) = even. this is not possible since this will not satisfy any of the cases(i) - (iv).
case (b): x = even, (y + z) = odd/even. this satisfies cases (i) and (ii).
so, since only case b is valid, x must be even. sufficient.
Statement 2:
y + xz = odd
case (c): y = odd, xz = even. this satisfies cases (ii) and (iv) where x could either odd or even. insufficient.
case (d): y = even, xz = odd. this satisfies case (iii), where x is odd.
so, since this statement doesn't provide any restriction on x, it is not sufficient.
Choose A.
-BM-
(i) x = even, y = even, z = odd
(ii) x = even, y = odd, z = odd
(iii) x = odd, y = even, z = odd
(iv) x = odd, y = odd, z = even
is x even?
Statement 1:
xy + xz = even
x(y + z) = even
case (a): x = odd, (y + z) = even. this is not possible since this will not satisfy any of the cases(i) - (iv).
case (b): x = even, (y + z) = odd/even. this satisfies cases (i) and (ii).
so, since only case b is valid, x must be even. sufficient.
Statement 2:
y + xz = odd
case (c): y = odd, xz = even. this satisfies cases (ii) and (iv) where x could either odd or even. insufficient.
case (d): y = even, xz = odd. this satisfies case (iii), where x is odd.
so, since this statement doesn't provide any restriction on x, it is not sufficient.
Choose A.
-BM-
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:Baldini 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
Could someone please explain how to get to the answer as quickly as possible?
Thanks
xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd
so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
-
- Legendary Member
- Posts: 940
- Joined: Tue Aug 26, 2008 3:22 am
- Thanked: 55 times
- Followed by:1 members
Cool solutionIan Stewart wrote:At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:Baldini 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
Could someone please explain how to get to the answer as quickly as possible?
Thanks
xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd
so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
- cubicle_bound_misfit
- Master | Next Rank: 500 Posts
- Posts: 246
- Joined: Mon May 19, 2008 7:34 am
- Location: Texaco Gas Station
- Thanked: 7 times
if this is of any help
from stmt 1,
xy +z +(x-1)z == even
odd + (x-1)z == even
hence (x-1)z is odd
hence x-1 odd and z odd
hence x even and z odd
putting back in the original condition
xy = even
z =odd
which satisfies the condition hence yes x is even, choose A/D
for stmt 2,
my reasoning is same as 'Bluementor'
hence answer A.
regards,
CBM
from stmt 1,
xy +z +(x-1)z == even
odd + (x-1)z == even
hence (x-1)z is odd
hence x-1 odd and z odd
hence x even and z odd
putting back in the original condition
xy = even
z =odd
which satisfies the condition hence yes x is even, choose A/D
for stmt 2,
my reasoning is same as 'Bluementor'
hence answer A.
regards,
CBM
Cubicle Bound Misfit
- bblast
- Legendary Member
- Posts: 1079
- Joined: Mon Dec 13, 2010 1:44 am
- Thanked: 118 times
- Followed by:33 members
- GMAT Score:710
Hi Ian,Ian Stewart wrote:At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:Baldini 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
Could someone please explain how to get to the answer as quickly as possible?
Thanks
xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd
so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
I remember you posted a similar solution to a similar GMAT club problem I posted here. Thanks for this approach on statement 1. Can we use similar method to deduce a result from statement 2 ? Or is testing cases is the only method out ?
Cheers !!
Quant 47-Striving for 50
Verbal 34-Striving for 40
My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_
Quant 47-Striving for 50
Verbal 34-Striving for 40
My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_
- czarczar
- Master | Next Rank: 500 Posts
- Posts: 106
- Joined: Fri Feb 05, 2010 9:39 am
- Thanked: 4 times
- Followed by:1 members
Perfect approach!Ian Stewart wrote:At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:Baldini 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
Could someone please explain how to get to the answer as quickly as possible?
Thanks
xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd
so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
- prateek_guy2004
- Master | Next Rank: 500 Posts
- Posts: 398
- Joined: Tue Jul 26, 2011 11:39 pm
- Location: India
- Thanked: 41 times
- Followed by:6 members
Statement 1 sufficient
1. xy + xz is an even integer
x(y+z) = even
2. y + xz is an odd integer Statement 2 Not sufficient.
Answer A
1. xy + xz is an even integer
x(y+z) = even
2. y + xz is an odd integer Statement 2 Not sufficient.
Answer A
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Statement 1:Baldini 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
Test whether it's possible that x is odd.
Let x=1.
Substituting x=1 into xy+z = ODD, we get:
y+z = ODD
Substituting x=1 into xy+xz = EVEN, we get:
y+z = EVEN
The 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 = ODD
Substituting x=1 into y+xz = ODD, we get:
y+z = ODD
Since 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 = ODD
Substituting x=2 into y+xz = ODD, we get:
y+2z = ODD
The 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.
The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3