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integer

by deepoe » Wed May 20, 2009 7:15 am
If x,y and z are positive integers such that x is a factor of y, and x is a multiple of z, wich 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 chose

X = 2
Y = 4
Z = 8

Did I made a wrong choice of picking numbers?

oa = B
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by dmateer25 » Wed May 20, 2009 7:21 am
2 is not a multiple of 8. That is your problem


multiple of 8 are 16, 24, 32, 40 ...


x is a factor of y, and x is a multiple of z


Let y = 8
let z =2
Let x = 4

Now 4 is a factor of 8 and 4 is a multiple of 2.

Plug the numbers in and you will see that B isn't an integer.
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by DanaJ » Wed May 20, 2009 7:28 am
Most of the time, picking numbers is a particularly risky strategy. It's best employed when trying to prove that something ISN'T true, like for instance if all numbers that are divisible by 3 are also divisible by 6. Pick 9 to prove that this statement is incorrect.
This is why I'll stick to that age old method of algebraically solving problems.

So you have that y = k*x - x is a factor of y
x = m*z - x is a multiple of z
Notice that since y = k*x = k*(mz) = km*z, so you have both y and x in terms of z:
x = k*z
y = km*z

I'd use these to eliminate choices:
a. x is divisible by z and z is also divisible by z. This means that the sum x + z will also be divisible by z, so this one is out.

b. y + z = km*z + z = z(km + 1)

(y + z)/x = z(km + 1)/kz = (km + 1)/k = km/k + 1/k = m + 1/k - this one is an integer only if 1/k is also an integer, which happens when k is either 1 or -1. As you can see, this is terribly restrictive, B is the correct answer. If you had this problem on the real thing, just pick B and go on to the next question. However, I will also analyze the other choices, just so you get a clear picture of things.

c. x is divisible by z and y is divisible by z, so their sum will also be divisible by z

d. since both x and y are divisible by z,xy will definitely be divisible by z

e. yz = km*z*z = km*(z^2)
x = k*z

yz/x = km*(z^2)/kz = zm - which will also be an integer.
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by Katrusya » Wed May 20, 2009 7:29 am
So, we have x as factor of y and z as factor of x.Therefore z is factor of y also.
Since x is factor of y, we can assume y=ax.
Since, z is factor of x, we can assume x=bz.
Hence, y=ax=abz.

Now all of the options we can present with a,b (those are constants) and z

A. ( x+z ) / z= (bz+z)/z=b+1-->integer
B. ( y+z )/ x = (abz+z)/bz=a+1/b--> not integer
C. ( x+y )/ z = (bz+abz)/z= b+ab--> integer
D. ( xy ) / z = (bz*abz)/z=abbz--> integer
E/ ( yz ) / x = (abz*z)/bz=az--> integer.

Therefore, B.
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by sanjeevbhatia3 » Wed May 20, 2009 9:28 am
we can also take following values:

x = 9
y = 45
z = 3

plug-in the value and you will see 'B' is the right answer

How much time this problem should take? As I took 20 seconds to solve this.
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Re: integer

by dumb.doofus » Wed May 20, 2009 9:40 am
deepoe wrote:If x,y and z are positive integers such that x is a factor of y, and x is a multiple of z, wich 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 chose

X = 2
Y = 4
Z = 8

Did I made a wrong choice of picking numbers?

oa = B
you picked the numbers correctly .. its just that you didnt assign them well.. else, its a 30 sec resolution problem..

When it says that X is a factor of Y and X is a multiple of Z, then in your example, X has to be 4, because only 4 satisfies the two criterias..
Y = 8 and Z = 2..
and Now it makes sense, why B cannot be an integer..
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