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consecutive odd integers

by prachi18oct » Wed Aug 20, 2014 6:41 am
If x, y, and z are consecutive odd integers, with x < y < z, then which of the following must be true?

x + y is even
is an integer
xz is even

I only
II only
III only
I and II only
I, II, and III

I eliminated option D as I considered -1,1,3 as one of the possibilities and hence -1+1 = 0 which is neutral.Please suggest if we should consider this possibility.
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by Brent@GMATPrepNow » Wed Aug 20, 2014 6:48 am
prachi18oct wrote:If x, y, and z are consecutive odd integers, with x < y < z, then which of the following must be true?

x + y is even
is an integer
xz is even

I only
II only
III only
I and II only
I, II, and III

I eliminated option D as I considered -1,1,3 as one of the possibilities and hence -1+1 = 0 which is neutral.Please suggest if we should consider this possibility.
You're missing part of the question (see above)

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by Brent@GMATPrepNow » Wed Aug 20, 2014 7:26 am
I found the original question. It reads as follows:
If x, y, and z are consecutive odd integers, with x < y < z, then which of the following MUST be true?

I. x + y is even
II. (x + z)/y is an integer
III. xz is even

A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III
The key word here is MUST

I. x + y is even
Since x and y are both ODD, x + y = ODD + ODD = EVEN
So, statement I is true

II. (x + z)/y is an integer
Must this be true?
Since x, y and z are consecutive odd integers, we know that y is 2 greater than x, and z is 4 greater than x.
So, we can write the following:
x = x
y = x + 2
z = x + 4

This means that (x + z)/y = (x + x + 4)/(x + 2)
= (2x + 4)/(x + 2)
= 2
Aha, so (x + z)/y will ALWAYS equal 2 (an integer)
So, statement II is true

III. xz is even
Since x and z are both ODD, xz = (ODD)(ODD) = ODD
So, statement III is NOT true

Answer: D

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by Brent@GMATPrepNow » Wed Aug 20, 2014 7:30 am
prachi18oct wrote:If x, y, and z are consecutive odd integers, with x < y < z, then which of the following must be true?

I. x + y is even
II. (x + z)/y is an integer
III. xz is even

A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III

I eliminated option D as I considered -1,1,3 as one of the possibilities and hence -1+1 = 0 which is neutral.Please suggest if we should consider this possibility.
Your mistake is highlighted above.

I believe you are confusing even/odd with positive/negative.
0 is neither positive nor negative.
But 0 is definitely even.

A number is "even" if that number can be written as the product of 2 and some integer.
Since we can write 0 as (2)(0), 0 is even.

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Brent
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by prachi18oct » Wed Aug 20, 2014 9:15 am
Yeah, my bad!! I got confused between even & positive/negative.
Thanks Brent!!
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by [email protected] » Wed Aug 20, 2014 11:39 am
Hi prachi18oct,

Roman Numeral questions are sometimes "built" to take longer to solve than typical Quant questions. However, they almost always involve Number Properties, which can be used to save time (if you know your Number Property rules).

From the onset of the question, we're told that X, Y and Z are CONSECUTIVE ODD INTEGERS. That type of vocabulary is a HUGE clue to think about Number Properties (re "patterns" in numbers). As Brent has shown in his explanation, you can quickly prove that Roman Numeral I is true and Roman Numeral III is false.

Be on the lookout for these patterns. Number Properties tend to show up all over the Quant section (especially in DS); knowing them will give you a great way to save time and pick up points.

GMAT assassins aren't born, they're made,
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by j_shreyans » Wed Aug 20, 2014 10:55 pm
Hi Brent ,

Can you pls explain why did you take y=x+2 if x=x it must be consecutive. I am bit confused in this.

Thanks

Shreyans
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by Matt@VeritasPrep » Wed Aug 20, 2014 11:10 pm
j_shreyans wrote:Hi Brent ,

Can you pls explain why did you take y=x+2 if x=x it must be consecutive. I am bit confused in this.

Thanks

Shreyans
Not to step on Brent's toes, but he did this because the numbers are consecutive ODD integers, which must differ by 2 (e.g. 1 and 3 or 3 and 5).
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by Brent@GMATPrepNow » Thu Aug 21, 2014 6:19 am
j_shreyans wrote:Hi Brent ,

Can you pls explain why did you take y=x+2 if x=x it must be consecutive. I am bit confused in this.

Thanks

Shreyans
Let's examine some consecutive ODD integers: 7, 9, 11
Notice that each integer is 2 greater than the one before it.
So, for example, 9 is 2 greater than 7, and 11 is 4 greater than 7
We can write:
7 = 7
9 = 7 + 2
11 = 7 + 4

Likewise, if x, y and z are consecutive ODD integers, then:
x = x
y = x + 2
z = x + 4

Once we know this, we can take (x + z)/y and replace y with x + 2 and replace z with x + 4 to get (x + x + 4)/(x + 2), which simplifies to (2x + 4)/(x + 2), which simplifies to 2

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Re: consecutive odd integers

by Scott@TargetTestPrep » Fri Feb 14, 2020 1:11 pm
prachi18oct wrote: ↑
Wed Aug 20, 2014 6:41 am
 If x, y, and z are consecutive odd integers, with x < y < z, then which of the following must be true?

x + y is even
is an integer
xz is even

I only
II only
III only
I and II only
I, II, and III

I eliminated option D as I considered -1,1,3 as one of the possibilities and hence -1+1 = 0 which is neutral.Please suggest if we should consider this possibility.
(Note that 0 is an even integer, so D can not be eliminated based on your statement.)

Since x, y, and z are consecutive odd integers, x + y = odd + odd = even, and xz = (odd)(odd) = odd. So statement I is true, and statement III is false.

Since x = y - 2 and z = y + 2, then (x + z)/y = (y - 2 + y + 2)/y = 2y/y = 2 is an integer. So statement II is true also.

Answer: D

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