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Suppose x, y, and z are positive integers such that xy + yz = 29 and x

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Intern
Joined: 25 Mar 2014
Posts: 31
Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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Updated on: 29 Aug 2014, 13:42
3
19
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Difficulty:

85% (hard)

Question Stats:

52% (01:33) correct 48% (01:16) wrong based on 414 sessions

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Suppose x, y, and z are positive integers such that xy + yz = 29 and xz + yz = 81. Which of the following variables has exactly one unique solution?

(i) x
(ii) y
(iii) z

A. none
B. ii only
C. iii only
D. i and ii only
E. ii and iii only

Originally posted by plaverbach on 29 Aug 2014, 12:19.
Last edited by Bunuel on 29 Aug 2014, 13:42, edited 1 time in total.
Edited the question
Intern
Joined: 25 Mar 2014
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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29 Aug 2014, 12:20
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How can one solve in under 3 min?

Official ansower above:
Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.

But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).

Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.

That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B.
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Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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30 Aug 2014, 07:06
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plaverbach wrote:
How can one solve in under 3 min?

Official ansower above:
Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.

But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).

Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.

That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B.

The way you have described should get you an answer in less than 2 mins.

To make the calculation a bit faster, recognise that in the second equation $$z * (x+1) = 81$$, 1 is already being added to x. So, it might be faster to substitute the value of x in terms of z.
Doing that, the equation becomes:

$$z * (29 - z + 1) = 81$$

$$30z - z^2 = 81$$

$$z^2 - 30z + 81 = 0$$

$$(z - 3) (z - 27) = 0$$

So, $$z = 3$$ or $$z = 27$$

Now, since z can take two values, without any further calculations, we know that x can take two values as well.

Hence, B is correct.

Hope that makes it clear.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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30 Aug 2014, 11:41
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Here's another approach:

Once we figure out that y=1, we use that to simplify the second equation to z(x+1) = 81. So what pairs of numbers make 81? 1*81, 3*27, or 9*9. Looking back at our first equation with y=1, we get x+z=29. So what do we know about x and z?

They add up to 29, and when we add 1 to x, they multiply to yield 81. The only pair of factors that works is 3 and 27. So we could have x=2 and y=27, or x=26 and y=3.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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03 Sep 2014, 21:22
1
xy + yz = 29

y(x+z) = 1 * 29 (29 is prime number; cannot be factorized in any other way)

y = 1 .............. (1)

x+z = 29

x = 29-z ........... (2)

xz + yz = 81

z(x+y) = 81

Placing values from (1) & (2)

z(29-z+1) = 81

$$30z - z^2 - 81 = 0$$

This would give 2 values for z (27, 3) which in turn would also yield 2 values for x; only y has a distinct value

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Suppose x,y, and z are positive integers such that  [#permalink]

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31 Jul 2015, 09:41
1
Suppose x,y, and z are positive integers such that xy+yz=29 and xz+yz=81 . Which of the following variables has exactly one unique solution?

(i) x

(ii) y

(iii) z

a)none
b)ii only
c)iii only
d)i and ii only
e)ii and iii only

Please give a kudos and help me unlock the awesome Gmatclub test
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Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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31 Jul 2015, 09:47
Gmat1008 wrote:
Suppose x,y, and z are positive integers such that xy+yz=29 and xz+yz=81 . Which of the following variables has exactly one unique solution?

(i) x

(ii) y

(iii) z

a)none
b)ii only
c)iii only
d)i and ii only
e)ii and iii only

Please give a kudos and help me unlock the awesome Gmatclub test

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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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30 Jun 2016, 23:20
1
1
plaverbach wrote:
Suppose x, y, and z are positive integers such that xy + yz = 29 and xz + yz = 81. Which of the following variables has exactly one unique solution?

(i) x
(ii) y
(iii) z

A. none
B. ii only
C. iii only
D. i and ii only
E. ii and iii only

Responding to a pm: The question relies on your knowledge of factors. Work step by step through the given information.

xy + yz = 29
(x + z)y = 29 (you need to immediately notice that 29 is prime)
So the two factors are 1 and 29 only. (x+z) cannot be 1 so y MUST be 1 and
(x + z) = 29

xz + yz = 81
(x + 1)z = 81
Factors of 81 -> 1, 3, 9, 27, 81.
The two factors cannot be 1 and 81 because (x+1) can be neither 1 nor 81 (since x + z is 29).
The two factors cannot be 9 and 9 because if z is 9(then x would be 20), x cannot be 8.
Then 3 and 27 must be the two factors. Now all we need to see is whether z can take both values 3 and 27.
If z is 3, x is 26 and total they add up to 29 - Works
If z is 27, x is 2 and total again adds up to 29 - Works

So x and z can take 2 values. Only y has a single value. Answer (B)
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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24 Mar 2017, 11:20
Found this question in the Veritas prep tests.
Like it, though it damaged my score.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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31 Jul 2017, 15:44
Here is a solution.. Answer is II or B
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x  [#permalink]

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30 Sep 2018, 22:02
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x &nbs [#permalink] 30 Sep 2018, 22:02
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