GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jan 2019, 03:05

### GMAT Club Daily Prep

#### 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

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

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.
• ### GMAT Club Tests are Free & Open for Martin Luther King Jr.'s Birthday!

January 21, 2019

January 21, 2019

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.

# If m and n are integers and x*y is defined as...

Author Message
TAGS:

### Hide Tags

Intern
Joined: 16 Dec 2018
Posts: 5
If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

29 Dec 2018, 01:20
00:00

Difficulty:

(N/A)

Question Stats:

46% (01:45) correct 54% (02:20) wrong based on 48 sessions

### HideShow timer Statistics

If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x · (– 1)^y, then what is the value of m*n?
(1) m*n = 2 + [(–1)^m · (–1)^n]
(2) (–1)^m· (–1)^n = 1
Manager
Joined: 01 May 2017
Posts: 69
Location: India
If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

29 Dec 2018, 02:05
1) $$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question
But given,
$$m*n = 2 + (-1)^m.(-1)^n$$
Which implies $$(-1)^m = (-1)^n = 1$$
Which in turn implies m and n are even
m*n = 3
Sufficient

2)$$(-1)^m.(-1)^n = 1$$
If both mand n are even or odd this is true
If both m & n are even m*n = 3
If both m & n are odd m* n = 1
Not sufficient

Hence correct option is A

Posted from my mobile device
Manager
Joined: 28 Jul 2016
Posts: 208
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
Re: If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

01 Jan 2019, 04:26
Mahimab wrote:
If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x · (– 1)^y, then what is the value of m*n?
(1) m*n = 2 + [(–1)^m · (–1)^n]
(2) (–1)^m· (–1)^n = 1

Can you edit the question to correct form :
_________________

Please give Kudos. Kudos encourage active discussions and help the community grow

Manager
Joined: 04 Oct 2017
Posts: 77
Re: If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

01 Jan 2019, 23:04
[quote="Dare Devil"]1) $$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question
But given,
$$m*n = 2 + (-1)^m.(-1)^n$$
Which implies $$(-1)^m = (-1)^n = 1$$
Which in turn implies m and n are even
m*n = 3
Sufficient

2)$$(-1)^m.(-1)^n = 1$$
If both mand n are even or odd this is true
If both m & n are even m*n = 3
If both m & n are odd m* n = 1
Not sufficient

Hence correct option is A

Hi,

Can you please explain how you got (-1) to the power of m = ( -1) to the power of n = 1?

Thank you.
Manager
Joined: 01 May 2017
Posts: 69
Location: India
Re: If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

01 Jan 2019, 23:33
Kezia9 wrote:
Dare Devil wrote:
1) $$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question
But given,
$$m*n = 2 + (-1)^m.(-1)^n$$
Which implies $$(-1)^m = (-1)^n = 1$$
Which in turn implies m and n are even
m*n = 3
Sufficient

2)$$(-1)^m.(-1)^n = 1$$
If both mand n are even or odd this is true
If both m & n are even m*n = 3
If both m & n are odd m* n = 1
Not sufficient

Hence correct option is A

Hi,

Can you please explain how you got (-1) to the power of m = ( -1) to the power of n = 1?

Thank you.

$$m*n = 2 + (-1)^m.(-1)^n$$

$$(-1)^m$$ OR $$(-1)^n$$ can be either 1 OR -1, since m, n are integers (may be positive or negative)
Since this is given,
$$m*n = 2 + (-1)^m.(-1)^n$$ ------ equation 1

$$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question [x,y replaced with m and n]
$$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ -------- equation 2
from equation 1 and equation 2
we know that $$(-1)^m + (-1)^n = 2$$
Since $$(-1)^(integer)$$ can only be 1 OR -1
possibilities of $$(-1)^m + (-1)^n$$ can be (0,-2,2)
Since from equation 1 and equation 2 we have deduced $$(-1)^m + (-1)^n$$ as 2, both m,n are even, which in turn implies $$(-1)^m$$ (which can be either 1 OR -1) is 1 and same for $$(-1)^n$$

I Hope this is clear
Manager
Joined: 04 Oct 2017
Posts: 77
Re: If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

02 Jan 2019, 00:10
Dare Devil wrote:
Kezia9 wrote:
Dare Devil wrote:
1) $$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question
But given,
$$m*n = 2 + (-1)^m.(-1)^n$$
Which implies $$(-1)^m = (-1)^n = 1$$
Which in turn implies m and n are even
m*n = 3
Sufficient

2)$$(-1)^m.(-1)^n = 1$$
If both mand n are even or odd this is true
If both m & n are even m*n = 3
If both m & n are odd m* n = 1
Not sufficient

Hence correct option is A

Hi,

Can you please explain how you got (-1) to the power of m = ( -1) to the power of n = 1?

Thank you.

$$m*n = 2 + (-1)^m.(-1)^n$$

$$(-1)^m$$ OR $$(-1)^n$$ can be either 1 OR -1, since m, n are integers (may be positive or negative)
Since this is given,
$$m*n = 2 + (-1)^m.(-1)^n$$ ------ equation 1

$$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question [x,y replaced with m and n]
$$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ -------- equation 2
from equation 1 and equation 2
we know that $$(-1)^m + (-1)^n = 2$$
Since $$(-1)^(integer)$$ can only be 1 OR -1
possibilities of $$(-1)^m + (-1)^n$$ can be (0,-2,2)
Since from equation 1 and equation 2 we have deduced $$(-1)^m + (-1)^n$$ as 2, both m,n are even, which in turn implies $$(-1)^m$$ (which can be either 1 OR -1) is 1 and same for $$(-1)^n$$

I Hope this is clear

Thank you for the prompt reply. I understood in part. m and n are even. I got this How can we conclude m*n = 3?Product of mn could be be multiple of 2.
Manager
Joined: 01 May 2017
Posts: 69
Location: India
If m and n are integers and x*y is defined as...  [#permalink]

### Show Tags

02 Jan 2019, 00:56
1
Kezia9 wrote:
Dare Devil wrote:
1) $$m*n = (-1)^m + (-1)^n + (-1)^m.(-1)^n$$ as per the question
But given,
$$m*n = 2 + (-1)^m.(-1)^n$$
Which implies $$(-1)^m = (-1)^n = 1$$
Which in turn implies m and n are even
m*n = 3
Sufficient

2)$$(-1)^m.(-1)^n = 1$$
If both mand n are even or odd this is true
If both m & n are even m*n = 3
If both m & n are odd m* n = 1
Not sufficient

Hence correct option is A

Hi,

Can you please explain how you got (-1) to the power of m = ( -1) to the power of n = 1?

Thank you.

m,n are integers

Case (a)
$$(-1)^m$$ = 1 if m is even
let the even integer be 4
(negative integer)^(even) = positive
$$(-1) ^4$$ = 1

Case (b)
$$if m is odd$$ (let it be 3)
$$(-1)^-3 = (-1/1)^3 =>(-1)^3$$ = -1
(negative integer)^(odd) = negative

Lets consider 3 cases,

Case(i)
m,n both are even
then from case (a)
$$(-1)^m + (-1)^n$$ = 1 + 1 = 2
which in turn implies
$$(-1)^m*(-1)^n$$ = 1*1 = 1 as m,n are even and from Case (a)

this implies $$m∗n=(−1)^m+(−1)^n+(−1)^m*(−1)^n$$ = 2+1 =3

case(ii)
m - even and n - odd OR m -odd and n - even
$$(-1)^m$$ = 1 (if m even, from case (a))
$$(-1)^n$$= -1 ( if n odd, from case (b))
So, $$(-1)^m + (-1)^n$$ = 1-1 = 0

Case(iii)
m,n both odd (from case (b))
($$-1)^m = (-1)^n$$ = -1 ($$(-1)^3$$ = -1)
So, $$(-1)^m + (-1)^n$$ = -1-1 = -2
If m and n are integers and x*y is defined as... &nbs [#permalink] 02 Jan 2019, 00:56
Display posts from previous: Sort by

# If m and n are integers and x*y is defined as...

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.