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If x, y, and z are integers, y + z = 13, and xz = 9, which o

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If x, y, and z are integers, y + z = 13, and xz = 9, which o [#permalink] New post 30 Aug 2018, 07:50
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If x, y, and z are integers, y + z = 13, and xz = 9, which of the following must be true?

(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x
[Reveal] Spoiler: OA

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Re: If x, y, and z are integers, y + z = 13, and xz = 9, which o [#permalink] New post 07 Sep 2018, 19:13
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sandy wrote:
If x, y, and z are integers, y + z = 13, and xz = 9, which of the following must be true?

(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x



xz = 9

Possible combinations:

3*3 = 9

1*9 = 9

9*1= 9

Now we can fint out the different values of y substituting the value of z.

y + z = 13

y + 3 = 13

y = 10

again,

y + z = 13

y + 1 = 13

y = 12

again,

y + z = 13

y + 9 = 13

y = 4.

In all cases y>3.

The best answer is D.
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Re: If x, y, and z are integers, y + z = 13, and xz = 9, which o [#permalink] New post 09 Sep 2018, 09:47
Expert's post
sandy wrote:
If x, y, and z are integers, y + z = 13, and xz = 9, which of the following must be true?

(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x


The most limiting fact is that x, y, and z are INTEGERS

So, if xz = 9, there are only 6 possible cases:
CASE A: x = 1 and z = 9
CASE B: x = -1 and z = -9
CASE C: x = 9 and z = 1
CASE D: x = -9 and z = -1
CASE E: x = 3 and z = 3
CASE F: x = -3 and z = -3

Since we also know that y + z = 13, we can add the corresponding y-value to each of the 6 possible CASES to get:

CASE A: x = 1, z = 9, and y = 4
CASE B: x = -1, z = -9, and y = 22
CASE C: x = 9, z = 1, and y = 12
CASE D: x = -9, z = -1, and y = 14
CASE E: x = 3, z = 3, and y = 10
CASE F: x = -3, z = -3, and y = 16

When we scan the answer choices, we can see that D must be true since all 6 possible y-values are greater than 3.

Answer: D

Cheers,
Brent
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Re: If x, y, and z are integers, y + z = 13, and xz = 9, which o [#permalink] New post 13 Sep 2018, 12:15
xz = 9
y + z = 13

xz = 9--> x or y can be -3/ +3/ -9/ +9/-1/+1 --> we also don't know which is x and which is z. But each can have 6 values

y + z = 13

10+3 = 16 - 3 = 12 + 1 = 14 - 1 = 13 (all equations come to 13)

so y is even

Out of the options:
(A) x is even --> no
(B) x = 3 --> no and yes / not suff
(C) y is odd --> no its even
(D) y > 3 --> YES in all cases
(E) z < x --> we don't know exact value of x or z, not suff

D is best
Re: If x, y, and z are integers, y + z = 13, and xz = 9, which o   [#permalink] 13 Sep 2018, 12:15
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If x, y, and z are integers, y + z = 13, and xz = 9, which o

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