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

Question

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

(1) |x| = |y+1|
(2) x^y = x! + |y|

sag · Answer

u can refer... math-absolute-value-modulus-86462.html for absolute values...

neoreaves · Answer

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

lprassanth · Answer

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.

toshio86 · Answer

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...

lprassanth · Answer

It would be great if either of Bunnel or Fig clarify on this point.

dixitraghav · Answer

hey anybody could explain this? Bunnel?

toshio86 · Answer

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?

bhandariavi · Answer

Simple correction!
0 raised to the 0 power is undefined. U cant say 0^0 = 1.

toshio86 · Answer

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

Pipp · Answer

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

bogos · Answer

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

onedayill · Answer

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

utfan2424 · Answer

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.

crack700 · Answer

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

mainhoon · Answer

Bunuel, Can you please settle the issue? Thanks.

PS what is 0^0?

arundas · Answer

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.

Bunuel · Answer

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
eq{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.

DeeptiM · Answer

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

gmatmachoman · Answer

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.

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

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