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Re: problem solving [#permalink]
08 Feb 2011, 03:20

Expert's post

2

This post was BOOKMARKED

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).

Answer: B.

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

Re: problem solving [#permalink]
08 Feb 2011, 03:36

1

This post received KUDOS

1

This post was BOOKMARKED

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 _________________

Re: 172. og qr. integer [#permalink]
10 Mar 2011, 23:36

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.

Re: 172. og qr. integer [#permalink]
10 Mar 2011, 23:50

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?

Re: If x, y and z are positive integers such that x is a factor [#permalink]
27 Apr 2015, 10:00

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