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Positive/Negative: If x, y, and z are positive integers, is

Expert replies
by II » Sun Apr 13, 2008 3:16 am
If x, y, and z are positive integers, is x-y odd ?

(1) x = z^2
(2) y = (z - 1)^2

Thanks.
Last edited by II on Mon May 05, 2008 1:41 am, edited 1 time in total.
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Source: — Data Sufficiency |

Re: Number Properties...

by Musiq » Sun Apr 13, 2008 5:14 am
II wrote:If x, y, and z are positive integeres, is x-y odd ?

(1) x = z^2
(2) y = (z - 1)^2

Thanks.
Interesting question II.

This is the very essence of Data Sufficiency. Within 10-20 seconds you should be able to eliminte A / B and D as possible answers.

Why?

Since the relationship being tested requires us to look for the outcome of X - Y, it becomes obvious that Statement 1 and Statement 2 by themselves are not going to pull it off (they only have X and Z, and Y and Z respectively.

If on test-da your time is running out (or even if it is not), something like this is really going to help you.

Now to decide between C and E:

Together:
X - Y = Z^2 - (Z-1)^2.................(This is the classic difference of squares rule)

Therfore, X- Y = ( Z - (Z-1)) ( Z+(Z+1))
which is then = (Z - Z +1) (2Z+1)
which simplifies as = (1) (2Z +1) = 2Z +1

The odd number kingdom is defined as 2(integer) + 1 OR 2(Integer) - 1.

Since Z is an integer, 2Z+1 MUST be odd. This answers the question with an ALWAYS YES, and hence sufficiency.

C is the answer.
For love, not money.
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by yalephd2007 » Sun Apr 13, 2008 8:19 am
I will go with C, too.
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by nikhilgmat31 » Wed Jun 17, 2015 12:14 am
It should be 2z-1

Answer is C
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by Brent@GMATPrepNow » Wed Jun 17, 2015 7:33 am
If x, y, and z are positive integers, is x-y odd?

1) x = z²
2) y = (z-1)²
Here's an algebraic approach:

Target question: Is x-y odd?

Given: x, y, and z are positive integers

Statement 1: x = z²
There's no information about y, so there's no way to determine whether or not x-y is odd.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y = (z-1)²
There's no information about x, so there's no way to determine whether or not x-y is odd.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x = z²
Statement 2: y = (z-1)²
Subtract equations to get: x-y = z² - (z-1)²
Expand to get: x-y = z² - [z² - 2z + 1]
Simplify to get: x-y = 2z - 1
Since z is a positive integer, we know that 2z is EVEN, which means 2z-1 is ODD.
If 2z-1 is ODD, we can conclude that x-y is definitely ODD
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Wed Jun 17, 2015 9:41 am
II wrote:If x, y, and z are positive integers, is x-y odd ?

(1) x = z^2
(2) y = (z - 1)^2
Clearly, neither statement alone is sufficient.

Statements combined:
Since x and y are both positive integers in terms of positive integer z -- and neither statement involves division -- we can determine whether x-y must be ODD by testing two cases:
z = EVEN and z = ODD.

Case 1: z=2
Here, x = 2² = 4 and y = (2-1)² = 1.
In this case, x-y = 4-1 = 3, which is ODD.

Case 2: z=3
Here, x = 3² = 9 and y = (3-1)² = 4.
In this case, x-y = 9-4 = 5, which is ODD.

Since x-y is ODD in both cases, the answer to the question stem is YES.
Thus, the two statements combined are SUFFICIENT.

The correct answer is C.
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by nikhilgmat31 » Fri Jun 26, 2015 2:54 am
Answer is C
2z-1 is always ODD
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by Matt@VeritasPrep » Mon Jun 29, 2015 4:29 pm
Another approach:

z² - (z-1)² =
z² - (z² - 2z + 1) =
2z - 1

For any integer z, 2z is even, so 2z - 1 is odd. We're set!
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