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# If x, y and z are positive integers such that x is a factor

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Manager
Joined: 04 Feb 2011
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If x, y and z are positive integers such that x is a factor [#permalink]

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08 Feb 2011, 03:34
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If x, y and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is not necessarily an integer?

A. (x+y)/z
B. (y+z)/x
C. (x+y)/z
D. (xy)/z
E. (yz)/x
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Apr 2012, 16:02, edited 1 time in total.
Edited the question and added the OA
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Joined: 02 Sep 2009
Posts: 39589

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08 Feb 2011, 04:20
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Lolaergasheva wrote:
If x, y and z are positive integers, such that x is a factor of y, and x is a multiple of z, which of the following is not necessarily an integer?

x+z/z
y+z/x
x+y/z
xy/z
yz/x

Given: $$z$$ goes into $$x$$ and $$x$$ goes into $$y$$. Note that it's not necessarily means that $$z<x<y$$, it means that $$z\leq{x}\leq{y}$$ (for example all three can be equal x=y=z=1);

Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: $$\frac{x+z}{z}$$ as $$z$$ goes into $$x$$ we can factor out it and reduce to get an integer result (or algebraically as $$x=zk$$ for some positive integer $$k$$ then $$\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer$$).

But in B. $$\frac{y+z}{x}$$ we can not be sure that we'll be able factor out $$x$$ from $$z$$ thus this option might not be an integer (for example x=y=4 and z=2).

Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.
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08 Feb 2011, 04:29
a little bit difficult to understand, how about to explain with real numbers?
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08 Feb 2011, 04:36
Lolaergasheva wrote:
a little bit difficult to understand, how about to explain with real numbers?

Try to plug y=8, x=4 and z=2 and solve.
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08 Feb 2011, 04:36
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From the statement;
z <= x <= y; such that;
x is a multiple of z and y is a multiple of x.
$$\therefore$$ y is a multiple of z.

1. (x+z)/z = x/z+1
x is a multiple of z; so x/z= integer;
integer+1 = integer. ALWAYS INTEGER.

2. (z+y)/x = z/x + y/x
y/x is always an integer because y is a multiple of x.
z/x is not necessarily an interger.
NOT ALWAYS AN INTEGER.

3. (x+y)/z = x/z+y/z
x/z - always an integer. x is a multiple of z
y/z - always an integer. y is a multiple of z
integer+integer=integer. ALWAYS INTEGER.

4. xy/z = y*(x/z)
x/z - integer - x is a multiple of z
y* integer = integer. Integer multiplied by integer is always integer. ALWAYS INTEGER

5. yz/x = z *(y/x)
y/x - integer - y is a multiple of x
z* integer = integer. Integer multiplied by integer is always integer. ALWAYS INTEGER

Ans: B

One may also substitute and see; Perhaps faster.
z=2
x=4
y=8
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11 Mar 2011, 00:12
If x,y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

A. x+z/z
B. y+z/x
C. x+y/z
D. xy/z
E. yz/x

i did as:
x=9
y=18
z=3
all options except B are integer so B is the answer. Is there any other good approach?
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Posts: 194
Re: 172. og qr. integer [#permalink]

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11 Mar 2011, 00:36
Baten80 wrote:
If x,y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

A. x+z/z
B. y+z/x
C. x+y/z
D. xy/z
E. yz/x

i did as:
x=9
y=18
z=3
all options except B are integer so B is the answer. Is there any other good approach?

x is factor of y so $$y = x*A$$ where A is another positive integer
x is a multiple of z, so $$x = B*z$$ where B is another positive integer and $$y=A*B*z$$

Now $$(x+z)/z = x/z + 1 = B+1$$ so integer
$$(y+z)/x = (A*B*z+z)/(B*z) = A + 1/B$$ which will not be an integer if B is an integer greater than 1, so Answer B

We can quickly see that C reduces to $$B + A*B$$, so integer, D is $$A*B^2$$, so integer and E is $$A*z$$, so integer again.

It looks elaborate but can be done in less than 90 seconds with pen and paper and would always give right answer, whereas trying to plug numbers may or may not work.
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Re: 172. og qr. integer [#permalink]

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11 Mar 2011, 00:50
1
KUDOS
modifying my approach a little bit here and using algebra instead of picking numbers.
Given
y/x = integer
x/z = integer

This implies -
y/z = integer

Now B stands out. y+z/x
y/x + z/x
z/x is the inverse of the x/z. The reversal of the relationship may not be TRUE always. Hence z/x may NOT be an integer !
Baten80 wrote:
If x,y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

A. x+z/z
B. y+z/x
C. x+y/z
D. xy/z
E. yz/x

i did as:
x=9
y=18
z=3
all options except B are integer so B is the answer. Is there any other good approach?
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Joined: 09 Sep 2013
Posts: 15915
Re: If x, y and z are positive integers such that x is a factor [#permalink]

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27 Apr 2015, 11:00
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Re: If x, y and z are positive integers such that x is a factor [#permalink]

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24 May 2015, 15:00
are these the right answer choices? A and C are same.
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Re: If x, y and z are positive integers such that x is a factor [#permalink]

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24 May 2015, 15:00
are these the right answer choices? A and C are same.
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Re: If x, y and z are positive integers such that x is a factor [#permalink]

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27 May 2016, 03:43
Lolaergasheva wrote:
If x, y and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is not necessarily an integer?

A. (x+y)/z
B. (y+z)/x
C. (x+y)/z
D. (xy)/z
E. (yz)/x

Dear Moderators,
Answer choice A and C are same. This is a question from OG. I believe answer option A must be (x+z)/z. Could you please rectify this?
Thank you,
Nalin

Given:
1) x is a factor of y, and
2) x is a multiple of z. In other words, z is a factor of x
This implies that z is also a factor of y.

A. (x+z)/z z is a factor of x. Hence, this will always be an integer
B. (y+z)/x x is a factor of y but we don't know whether x is a factor of z or not. Hence, this could be or could not be an integer
C. (x+y)/z z is a factor of both x and y. Hence, this will always be an integer
D. (xy)/z z is a factor of both x and y. Hence, this will always be an integer
E. (yz)/xx is a factor of y. Hence, this will always be an integer

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Joined: 18 Sep 2016
Posts: 1
Re: If x, y and z are positive integers such that x is a factor [#permalink]

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05 Feb 2017, 14:17
Bunuel wrote:
Lolaergasheva wrote:
If x, y and z are positive integers, such that x is a factor of y, and x is a multiple of z, which of the following is not necessarily an integer?

x+z/z
y+z/x
x+y/z
xy/z
yz/x

Given: $$z$$ goes into $$x$$ and $$x$$ goes into $$y$$. Note that it's not necessarily means that $$z<x<y$$, it means that $$z\leq{x}\leq{y}$$ (for example all three can be equal x=y=z=1);

Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: $$\frac{x+z}{z}$$ as $$z$$ goes into $$x$$ we can factor out it and reduce to get an integer result (or algebraically as $$x=zk$$ for some positive integer $$k$$ then $$\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer$$).

But in B. $$\frac{y+z}{x}$$ we can not be sure that we'll be able factor out $$x$$ from $$z$$ thus this option might not be an integer (for example x=y=4 and z=2).

Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.

Hi, thanks for the answer. Do I understand it right that we have to watch for the wording "not necessarily"? Because if I choose for x=y=z = 1 I will receive an integer for answer B) as well.
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Joined: 02 Sep 2009
Posts: 39589
Re: If x, y and z are positive integers such that x is a factor [#permalink]

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06 Feb 2017, 07:31
Sergej111 wrote:
Bunuel wrote:
Lolaergasheva wrote:
If x, y and z are positive integers, such that x is a factor of y, and x is a multiple of z, which of the following is not necessarily an integer?

x+z/z
y+z/x
x+y/z
xy/z
yz/x

Given: $$z$$ goes into $$x$$ and $$x$$ goes into $$y$$. Note that it's not necessarily means that $$z<x<y$$, it means that $$z\leq{x}\leq{y}$$ (for example all three can be equal x=y=z=1);

Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: $$\frac{x+z}{z}$$ as $$z$$ goes into $$x$$ we can factor out it and reduce to get an integer result (or algebraically as $$x=zk$$ for some positive integer $$k$$ then $$\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer$$).

But in B. $$\frac{y+z}{x}$$ we can not be sure that we'll be able factor out $$x$$ from $$z$$ thus this option might not be an integer (for example x=y=4 and z=2).

Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.

Hi, thanks for the answer. Do I understand it right that we have to watch for the wording "not necessarily"? Because if I choose for x=y=z = 1 I will receive an integer for answer B) as well.

Yes. The question basically asks: which of the following is not ALWAYS an integer?
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Re: If x, y and z are positive integers such that x is a factor [#permalink]

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06 Feb 2017, 22:57
Lolaergasheva wrote:
a little bit difficult to understand, how about to explain with real numbers?

take x = 10 , y= 40 , z= 5. hence satisfies the above question. substitute and find as per the options. B
Joined: 30 May 2015
Posts: 26
Re: If x, y and z are positive integers such that x is a factor [#permalink]

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11 Feb 2017, 14:23
suppose
y = x*a ..... eq1(as x is a factor of y ; a is some integer here)
and
x = z*b ..... eq2 (as x is a multiple of z ; a is some integer here))

A. (x+y)/z
after putting the values from above equations
(x + x*a)/(x/b) => (1+a)b (It will be integer)

B. (y+z)/x
=> ((x*a + (x/b))/x) => (a + 1/b) (It is not necessary to be integer) Answer

C. (x+y)/z
=> ((x + x*a)/(x/b)) => (1 +a)b => It will be integer

D. (xy)/z => It will be integer after putting z's value in terms of x

E. (yz)/x => => It will be integer after putting y's value in terms of x here
Manager
Joined: 26 Mar 2017
Posts: 147
Re: If x, y and z are positive integers such that x is a factor [#permalink]

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14 May 2017, 07:03
nalinnair wrote:
Lolaergasheva wrote:
If x, y and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is not necessarily an integer?

A. (x+y)/z
B. (y+z)/x
C. (x+y)/z
D. (xy)/z
E. (yz)/x

Dear Moderators,
Answer choice A and C are same. This is a question from OG. I believe answer option A must be (x+z)/z. Could you please rectify this?
Thank you,
Nalin

Given:
1) x is a factor of y, and
2) x is a multiple of z. In other words, z is a factor of x
This implies that z is also a factor of y.

A. (x+z)/z z is a factor of x. Hence, this will always be an integer
B. (y+z)/x x is a factor of y but we don't know whether x is a factor of z or not. Hence, this could be or could not be an integer
C. (x+y)/z z is a factor of both x and y. Hence, this will always be an integer
D. (xy)/z z is a factor of both x and y. Hence, this will always be an integer
E. (yz)/xx is a factor of y. Hence, this will always be an integer

yes, this is the best approach imo.

I couldn't remember the rules so i just picked 2, 4 and 8
Re: If x, y and z are positive integers such that x is a factor   [#permalink] 14 May 2017, 07:03
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