GMAT PREP ?? (Multiples and Factors)

[dferm]

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

[codesnooker]

IMO = B

I consider, the options provided by you are as follows:-

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

FIRST APPROACH

Lets assume X = 9 (Don't take prime number, as X is multiple of Z)

so Z could be equal to 3

and as X is factor of Y, so we can take any value of Y that is completely divisible by 9.

Lets say Y = 45

so X = 9, Y = 45 and Z = 3

Now plug these values one by in each equation. Only for choice (B) the results comes into fraction. Hence (B) is the right answer.

SECOND APPROACH

A. (x+z)/z

For A, x is multiple of z. So we add one more z to x and then divide by z, it would result into the integer. (For example, if z = 2 and x = 6, so if add 2 to 6 and divide by 2, it would result into the integer. Simple thing: Its comes under the same table.)

B. (y+z)/x

For B, x is the factor of y but not of z. So, if we add z to y and try to divide it by x, then obviously it would result z as the reminder.

C. (x+y)/z

For C, x is multiple of z. It means z is factor of x. Its already given that x is factor of y. So by these two facts, its clear that z is also factor of y. So, when we add two number which are having common factor, then dividing by common factor would result into an integer.

D. (xy)/z

For D, same reason as I have stated for C. (only the operator symbol is changed which unaffected by the fact stated in the option C)

E. (yz)/x

For E, as x is factor of y, so when we divide y by x, it would result an integer. And then multiplying an integer with another integer would result an integer.


Choice is yours. Which ever you find easiest can blindly follow that.

[VP_RedSoxFan]

Since the question is which is NOT necessarily an integer, as soon as, by whatever method, you found one that leave a remainder. Move on. There will only be one answer.

In this case, as soon as the numbers you've plugged (I used Y=12, X=4, Z=2) leave a remainder (B), move on to the next question. You don't need to prove out all of the other answers. I think this is an important time management technique.

[codesnooker]

I think this is an important time management technique.

Absolutely!!! I have mentioned all the solution, so that if the correct answer is placed at E, then one should be able to reject other choices.

Thanks for adding the point.

[f00kie]

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

Use equations.

x is a factor of y means y = ax.
x is a multiple of z means x = bz.

a and b are both positive integers.

Then, sub it into the above equations, until you have something awkward.
In this case, I hit it at B, since it produced:

(ab + 1) / a which is clearly not necessarily an integer.

[maihuna]

I am confused with option B here:

(y+z)/x = (kx + x/m)/x = k+ 1/m assuming: y=kx and x = mz

sow how k+1/m will be an integer only, for m=2 it will noy be right? experts, clarify?????

[arora007]

Great post!

[GMATGuruNY]

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

Our goal is to find the answer choice that does not have to be an integer.
The conditions are that z is a factor of x and and that x is a factor of y.
Since z is a factor both of x and of y, the answer choices in which z is in the denominator (A, C and D) are unlikely to be correct.

Let's try to show that B does not have to be an integer.

Plug in z=1, x=2, y=4.

(y+z)/x = (4+1)/2 = 5/2. Not an integer.

The correct answer is B.

[Everest]

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

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


solution) Let's plug in some numbers to solve these type of questions.

X is factor of Y => Y is multiple of X
X is a multiple of z

+> Y is multiple of X and X is multiple of Z

let's assign numbers Z = 2, X = 4 and Y = 8

plug in to options A, B, C, D and E

Only choice B results in a fraction

so, IMO is B

[aleph777]

I thought I'd add my two cents to show another way of thinking about this one using number theory.

There are a few steadfast rules to remember regarding any numbers:

1. If A is a factor of B and B is a factor of C, then A is also a factor of C. Numerically, if 2 is a factor of 4, and 4 is a factor of 8, then 2 is also a factor of 8.

2. If you add or subtract multiples of N, you get a multiple of N.

3. If you add or subtract a multiple of N and a non-multiple of N, you never get a multiple of N.

4. If you add or subtract two non-multiples of N, you COULD get a multiple of N.

--------------------

Since we know that x, y, and z, all share z as a factor, we can then apply these rules.

A. x+z/z
x and z both share z as a factor. Therefore, their sum is divisible by z.

B. y+z/x
x is a factor of y, but not of z. If you add or subtract a multiple of x and a non-multiple of x, you never get a multiple of x. Therefore, their sum is not divisible by x.

C. x+y/z
z is a factor of both x and y. Therefore, their sum is divisible by z.

D. xy/z
z is a factor of both x and y, therefore their product is divisible by z.

E. yz/x
Although x is not a factor of z, it is a factor of y. And therefore the product of y and z is divisible by x.

Hope that helps!