GMAT PREP ?? (Multiples and Factors)

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.

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.

VP_RedSoxFan wrote: 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.

dferm 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

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.

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?????

Great post!

dferm 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

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.

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

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.

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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!