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If x and y are integers, is |x| > |y|? 1) |x| = |y+1| 2)

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If x and y are integers, is |x| > |y|? 1) |x| = |y+1| 2) [#permalink] New post 05 May 2010, 06:28
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D
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32% (02:46) correct 68% (01:44) wrong based on 63 sessions
If x and y are integers, is |x| > |y|?

(1) |x| = |y+1|
(2) x^y = x! + |y|
[Reveal] Spoiler: OA
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Re: Ineq + absolute values [#permalink] New post 11 May 2010, 22:21
u can refer...

math-absolute-value-modulus-86462.html

for absolute values...
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Re: Ineq + absolute values [#permalink] New post 12 May 2010, 06:33
1) this is not enough ....lets look at two possibilities

x = 0 y = -1
in this case |x| > |y|
x = 3 y = 2
in this case |x| < |y|

So insufficient


2) x! is always positive so adding this to |y| will increase y thus x > y

so |x| > |y|

Sufficient

So IMO answer should be B
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Re: Ineq + absolute values [#permalink] New post 14 May 2010, 21:40
Quote:
2) x! is always positive [highlight]so adding this to |y| will increase y thus x > y
so |x| > |y|[/highlight]
Sufficient
So IMO answer should be B

I'm curious how you can say which one of x or y is greater. Does this statement not just say that x,y>0. So should the answer not be (C).
Please correct me if/where i'm wrong.
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Re: Ineq + absolute values [#permalink] New post 14 May 2010, 22:44
Sorry I don't understand why the 2nd stmt is sufficient...

If we consider x^y=x! + |y| and we plug in y=0, we get two solutions:

x=0 (since 0^0=1 and 0!=1)
x=1 (since 1^0=1 and 1!=1)

So we have y=0, x=0 >>>> |x|=|y|
but we also have y=0, x=1 >>>> |x|>|y|

(if you try with y=2, you will get x=2, so y=x again)

So stmt 2 should be insufficient...
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Re: Ineq + absolute values [#permalink] New post 15 May 2010, 06:14
It would be great if either of Bunnel or Fig clarify on this point.
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Re: Ineq + absolute values [#permalink] New post 15 May 2010, 07:00
hey anybody could explain this? Bunnel?
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Re: Ineq + absolute values [#permalink] New post 15 May 2010, 08:33
Just to give it a try, while waiting for Bunuel... :wink:

stmt 1

If x=0, y=-1, stmt 1 is satisfied, and |x|<|y|
If x=1, y=0, stmt 1 is satisfied, and |x|>|y|

not sufficient

stmt 2

If we consider x^y=x! + |y| and we plug in y=0, we get two solutions:

x=0 (since 0^0=1 and 0!=1)
x=1 (since 1^0=1 and 1!=1)

So we have y=0, x=0 >>>> |x|=|y|
but we also have y=0, x=1 >>>> |x|>|y|

(if you try with y=2, you will get x=2, so |x|=|y| again)

not sufficient

stmt 1 + stmt 2

The only acceptable solution that satisfies both stmt 1 and 2 is
y=0, x=1

If you try to plug in stmt 2 other values for x and y that satisfy stmt 1 (eg, x=2, y=1; x=3, y=2; x=5, y=4), you'll see that there are no other acceptable solutions.

So, if y=0 and x=1, |x|>|y|

Answer C (I hope so!)

Any opinions?
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Re: Ineq + absolute values [#permalink] New post 15 May 2010, 10:38
Simple correction!
0 raised to the 0 power is undefined. U cant say 0^0 = 1.
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Re: Ineq + absolute values [#permalink] New post 15 May 2010, 18:20
it looks like 0^0=1, at least according to most of the sources...
Have a look here:

gmatclub dot com/forum/0-raised-to-9060 dot html (sorry I'm a new member and I can't post links yet!)

but it also says we shouldn't need 0^0 for the GMAT :?
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Re: Ineq + absolute values [#permalink] New post 30 Jun 2010, 22:54
ii. x^y = x! + |y|


x! and |y| are both integers, so x! + |y| is also integer, therefore x^y must be integer, which means y must be greater or equal zero so as not to make x^y a fraction. Now knowing that x,y are integers greater than or equal zero, we need to find the value of x,y for which the equation is true. I can only find x=2 y=2 that satisfy the equation, which gives me a B
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Re: Ineq + absolute values [#permalink] New post 27 Jul 2010, 05:11
My answer is E.

Statement1:
First, recall that |a+b|<=|a|+|b|, therefore
|x|=|y+1|<=|y|+1
meaning that |x| is either equal to |y|+1 or inferior to |y|+1
then NOT SUFFICIENT

Statement 2:
as x! is used, hence x>=0
if x=0, then y=0, since 0^0 is not accepted, then x>0
if y=0, then x^0=x!+0, or x=1, therefore |x|>|y|
if y=1, then x^1=x!+1, there is no solution of x
if y=2, then x^2=x!+2, or x=2, therefore |x|=|y|
then NOT SUFFICIENT
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Re: Ineq + absolute values [#permalink] New post 27 Jul 2010, 06:59
I'll go with C.

1 is clearly insufficient...

2: Try substituting values of x =0 and y =1 and vice versa.
0 = 1
1 = 1.

Not sufficient.

Together:

The only acceptable solution that satisfies both stmt 1 and 2 is
y=0, x=1

C
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Re: Ineq + absolute values [#permalink] New post 28 Jul 2010, 08:52
I think it is C.

(1) If Y is negative, X<Y. If Y is positive or zero, X>Y. (but at least we know they're both integers)
(2) We know X! is always positive and b/c both X&Y are both integers, Y cannot be negative (since anything raised to a negative power, besides 1 or zero, is a non-integer. We can also know that X,Y cannot be 0,1 (b/c you get 0 on the left and 1 on the right), X,Y cannot be 1,1. X,Y CAN be 1,0. We found out that Y & X must both be positive or X,Y is 1,0.
We can't say for sure which is larger. NOT SUFFICIENT

Using both we see that b/c Y is positive or zero, X>Y. So I choose C. Let me know if someone catches something I missed - first post.
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Re: Ineq + absolute values [#permalink] New post 28 Jul 2010, 18:15
I will go with C.
As per (1) |x|=|y+1|. If y is positive or zero then |x|>|y|
If y is negative then |x|<|y|. So not sufficient.

As per 2
x! is used so x>=0
x!+|y|>=0 and is an integer.So x^y can't be a fraction which means y>=0
But x=1 y=0 and x=2 y=2 produces different results. So not sufficient

Combining 1 and 2
As per 2, x and y both have to be positive So 1 can be re-written as x= y+1
Which means x > y and |x|>|y| as both are positive. So sufficient
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Re: Ineq + absolute values [#permalink] New post 07 Aug 2010, 08:16
Bunuel, Can you please settle the issue? Thanks.

PS what is 0^0?
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Re: Ineq + absolute values [#permalink] New post 08 Aug 2010, 20:51
I was surprises when I read 0 ^ 0 = 1. So I went and check with my best Friend Google. Typed 0 ^ 0 and google returned 1. So I think 0 ^ 0 = 1.
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Re: Ineq + absolute values [#permalink] New post 09 Aug 2010, 02:04
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I was reluctant to reply to this question for a long time because I don't like it all.

First of all: 0^0, in some sources equals to 1, some mathematicians say it's undefined. Anyway you won't need this for GMAT as the case of 0^0 is not tested on the GMAT.

Second: GMAT would give some constraints for unknowns in the stem (or at leas in second statement), for example:

As there is x! in statement (2) then in the stem (or at leas in second statement) GMAT most likely would specify that x\geq{0} as factorial of negative number is undefined;

Also as there is x^y in statement (2) then in the stem (or at leas in second statement) GMAT most likely would also specify that x and y can not be zero simultaneously as 0^0 is not tested on the GMAT.

So either we would have A. x\geq{0} and y\neq{0} OR B. x>0.

If we place these constraint in the stem the question will be completely changed (no need for |x| in the stem and statement 1.) so we should place either of them them in statement 2.

So the question could be for example:

If x and y are integers, is |x|>|y|?

(1) |x| = |y+1|. Clearly insufficient.

(2) x>{0} and x^y = x! + |y| --> if x=2 and y=2 then answer is NO but if x=1 and y=0 then answer is YES. Not sufficient.

(1)+(2) Only one solution x=1 and y=0. Sufficient.

Answer: C.
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Algebra - Tough! [#permalink] New post 26 Aug 2011, 11:11
If x and y are integers, is |x| > |y|?
(1) |x| = |y + 1|
(2) x^y = x! + |y|
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Re: Algebra - Tough! [#permalink] New post 26 Aug 2011, 11:32
DeeptiM wrote:
If x and y are integers, is |x| > |y|?
(1) |x| = |y + 1|
(2) x^y = x! + |y|


St 1 : case 1 X= -9 , y = -10 ( where |X| < |Y|)

|-9| = |-10 +1|

----> 9 = 9

case 2: where |X| > |Y|

X = -3 , Y = 2

Insufficient.

St 2 : X = 2, Y = 2 & X = 1, Y = 0

Inconsistent results. Insufficient.

Combining st 1 & st 2;

use X = 1, Y = 0 ( using both the statements together)

we see |X| > |Y|.

Pick C.
Re: Algebra - Tough!   [#permalink] 26 Aug 2011, 11:32
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