Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 14 Apr 2010
Posts: 222

If x and y are integers, is x > y? 1) x = y+1 2) [#permalink]
Show Tags
05 May 2010, 07:28
Question Stats:
33% (02:37) correct
68% (01:42) wrong based on 80 sessions
HideShow timer Statistics
If x and y are integers, is x > y? (1) x = y+1 (2) x^y = x! + y
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 20 Apr 2010
Posts: 153
Location: I N D I A

Re: Ineq + absolute values [#permalink]
Show Tags
11 May 2010, 23:21



Manager
Joined: 09 Apr 2010
Posts: 79

Re: Ineq + absolute values [#permalink]
Show Tags
12 May 2010, 07:33
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



Intern
Joined: 07 Apr 2010
Posts: 23

Re: Ineq + absolute values [#permalink]
Show Tags
14 May 2010, 22:40
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.
_________________
In a Normal Distribution, only the Average 'Stand Out'



Intern
Joined: 14 May 2010
Posts: 33
Schools: CBS

Re: Ineq + absolute values [#permalink]
Show Tags
14 May 2010, 23:44
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...



Intern
Joined: 07 Apr 2010
Posts: 23

Re: Ineq + absolute values [#permalink]
Show Tags
15 May 2010, 07:14
It would be great if either of Bunnel or Fig clarify on this point.
_________________
In a Normal Distribution, only the Average 'Stand Out'



Manager
Joined: 16 Feb 2010
Posts: 186

Re: Ineq + absolute values [#permalink]
Show Tags
15 May 2010, 08:00
hey anybody could explain this? Bunnel?



Intern
Joined: 14 May 2010
Posts: 33
Schools: CBS

Re: Ineq + absolute values [#permalink]
Show Tags
15 May 2010, 09:33
Just to give it a try, while waiting for Bunuel... 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?



Manager
Joined: 04 Apr 2010
Posts: 162

Re: Ineq + absolute values [#permalink]
Show Tags
15 May 2010, 11:38
Simple correction! 0 raised to the 0 power is undefined. U cant say 0^0 = 1.
_________________
Consider me giving KUDOS, if you find my post helpful. If at first you don't succeed, you're running about average. ~Anonymous



Intern
Joined: 14 May 2010
Posts: 33
Schools: CBS

Re: Ineq + absolute values [#permalink]
Show Tags
15 May 2010, 19:20
it looks like 0^0=1, at least according to most of the sources... Have a look here: gmatclub dot com/forum/0raisedto9060 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



Intern
Joined: 19 Jun 2010
Posts: 27

Re: Ineq + absolute values [#permalink]
Show Tags
30 Jun 2010, 23:54
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



Intern
Joined: 03 Mar 2010
Posts: 40

Re: Ineq + absolute values [#permalink]
Show Tags
27 Jul 2010, 06:11
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
_________________
Hardworkingly, you like my post, so kudos me.



Senior Manager
Joined: 25 Feb 2010
Posts: 461

Re: Ineq + absolute values [#permalink]
Show Tags
27 Jul 2010, 07:59
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
_________________
GGG (Gym / GMAT / Girl)  Be Serious
Its your duty to post OA afterwards; some one must be waiting for that...



Intern
Joined: 05 Nov 2009
Posts: 32

Re: Ineq + absolute values [#permalink]
Show Tags
28 Jul 2010, 09:52
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 noninteger. 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.



Manager
Joined: 20 Mar 2010
Posts: 84

Re: Ineq + absolute values [#permalink]
Show Tags
28 Jul 2010, 19:15
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 rewritten as x= y+1 Which means x > y and x>y as both are positive. So sufficient
_________________
___________________________________ Please give me kudos if you like my post



Director
Status: Apply  Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 685
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

Re: Ineq + absolute values [#permalink]
Show Tags
07 Aug 2010, 09:16
Bunuel, Can you please settle the issue? Thanks. PS what is 0^0?
_________________
Consider kudos, they are good for health



Manager
Status: ISB, Hyderabad
Joined: 25 Jul 2010
Posts: 173
WE 1: 4 years Software Product Development
WE 2: 3 years ERP Consulting

Re: Ineq + absolute values [#permalink]
Show Tags
08 Aug 2010, 21:51
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.
_________________
AD



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Ineq + absolute values [#permalink]
Show Tags
09 Aug 2010, 03:04
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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 16 Feb 2011
Posts: 258

Algebra  Tough! [#permalink]
Show Tags
26 Aug 2011, 12:11
If x and y are integers, is x > y? (1) x = y + 1 (2) x^y = x! + y



Intern
Joined: 28 Aug 2010
Posts: 3

Re: Algebra  Tough! [#permalink]
Show Tags
26 Aug 2011, 12:32
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, 12:32







