Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.