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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'

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.

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

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 _________________

(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.

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 _________________

___________________________________ Please give me kudos if you like my post

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. _________________

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.

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