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# If x < y < z and y-x > 5, where x is an even integer and y a

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If x < y < z and y-x > 5, where x is an even integer and y a [#permalink]

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07 Apr 2011, 05:26
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If x < y < z and y - x > 5, where x is an even integer and y and z are odd integers, what is the least possible value of z - x?

A. 6
B. 7
C. 8
D. 9
E. 10

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-z-and-y-x-5-where-x-is-an-even-integer-and-y-129754.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Oct 2013, 02:07, edited 1 time in total.
Renamed the topic and edited the question.
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Re: number properties [#permalink]

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07 Apr 2011, 05:41
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Eliminating the choices since answer is odd. Only B and D left.

B Vs D
Let x is 0. The smallest value of y is 7. The next odd is 9. Hence z = 9.
z - x = 9

Acer86 wrote:
If x < y < z and y - x > 5, where x is an even integer and y and z are odd integers, what is the least possible value of z – x ?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

The answer i am getting is 7..thought original answer is something else...can someone help me out

Last edited by gmat1220 on 07 Apr 2011, 06:39, edited 1 time in total.
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Re: number properties [#permalink]

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07 Apr 2011, 05:44
Acer86 wrote:
If x < y < z and y - x > 5, where x is an even integer and y and z are odd integers, what is the least possible value of z – x ?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

The answer i am getting is 7..thought original answer is something else...can someone help me out

x<y<z
To find the least possible value for z-x; we need to find the values for z and x that can be closest to each other. If x is some even number, then what could be minimum possible odd z.

If x is some even number
y-x>5; y>x+5; minimum value for y=x+5+2=x+7[Note: x+5 is as even+odd=odd and nearest odd greater than x+5 is x+5+2]
Minimum value for z=y+2=x+7+2=x+9 [Note: z=y+2 because both z and y are odd. Difference between two odd numbers is 2]

Thus,
z-x = x+9-x= 9

Ans: "D"
********************************

Or, just use substitution:

Pick x as any even number
x=10
y> x+5 and Y is ODD
y> 15 and is ODD
Closest odd from 15 which is greater than 15 is 15+2=17
z is odd but more than y
Thus, z=y+2=17+2=19

z-x=19-10=9

Ans: "D"
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Re: number properties [#permalink]

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07 Apr 2011, 05:55
z - x is least when z is least and x is max

y > 5 + x

For y to be minimum, x = 0 (lowest even), and y > 5, so y = 7

hence z = 9, because z is odd too

so 9 - 0 = 9

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Director
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Kudos [?]: 346 [0], given: 123

Re: number properties [#permalink]

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07 Apr 2011, 06:22
z - x is least when z is least and x is max ---> I think this is not true. The least value of z is -Infinity and max value of x is +Infinity. So by this rule the min value is -Infinity

So the assumption should be z and x are closeby for z-x to be minimum.
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Re: number properties [#permalink]

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10 Oct 2013, 10:02
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Re: If x < y < z and y-x > 5, where x is an even integer and y a [#permalink]

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11 Oct 2013, 02:08
If x < y < z and y - x > 5, where x is an even integer and y and z are odd integers, what is the least possible value of z - x?

A. 6
B. 7
C. 8
D. 9
E. 10

We want to minimize $$z-x$$, so we need to maximize $$x$$.

Say $$z=11=odd$$, then max value of $$y$$ will be 9 (as $$y$$ is also odd). Now, since $$y-5>x$$ --> $$9-5>x$$ --> $$4>x$$, then max value of $$x$$ is 2 (as $$x$$ is even).

Hence, the least possible value of $$z-x$$ is 11-2=9.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-z-and-y-x-5-where-x-is-an-even-integer-and-y-129754.html
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Re: If x < y < z and y-x > 5, where x is an even integer and y a   [#permalink] 11 Oct 2013, 02:08
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# If x < y < z and y-x > 5, where x is an even integer and y a

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