Author 
Message 
Intern
Joined: 16 Dec 2018
Posts: 6

If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
29 Dec 2018, 02:20
Question Stats:
42% (02:13) correct 58% (02:34) wrong based on 79 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) \((–1)^m*(–1)^n = 1\)
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 01 May 2017
Posts: 83
Location: India

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
29 Dec 2018, 03:05
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



Director
Joined: 28 Jul 2016
Posts: 675
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
01 Jan 2019, 05:26
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 : m*n instead of x*y



Manager
Joined: 04 Oct 2017
Posts: 73

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
02 Jan 2019, 00:04
[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: 83
Location: India

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
02 Jan 2019, 00:33
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 2from equation 1 and equation 2 we know that \((1)^m + (1)^n = 2\) Since \((1)^(integer)\) can only be 1 OR 1possibilities 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: 73

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
02 Jan 2019, 01:10
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 2from equation 1 and equation 2 we know that \((1)^m + (1)^n = 2\) Since \((1)^(integer)\) can only be 1 OR 1possibilities 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: 83
Location: India

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
02 Jan 2019, 01:56
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) = negativeLets consider 3 cases, Case(i) m,n both are even then from case (a) \((1)^m + (1)^n\) = 1 + 1 = 2which 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, \((1)^m + (1)^n\) = 11 = 0Case(iii) m,n both odd (from case (b)) (\(1)^m = (1)^n\) = 1 (\((1)^3\) = 1) So, \((1)^m + (1)^n\) = 11 = 2



Manager
Status: wake up with a purpose
Joined: 24 Feb 2017
Posts: 154
Concentration: Accounting, Entrepreneurship

Re: If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
13 Mar 2019, 22:59
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 Bunuel Would you please explain this question? I don't understand explanations above. Posted from my mobile device
_________________
If people are NOT laughing at your GOALS, your goals are SMALL.



Director
Joined: 24 Oct 2016
Posts: 586
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38

If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
Show Tags
17 Mar 2019, 05:57
globaldesi wrote: 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 : m*n instead of x*y BunuelI believe there's a typo in this Q. I think it should be " m*n is defined as" instead of " x*y is defined as". Can you please correct it? Thanks!




If m and n are integers and x*y is defined as (–1)^x + (–1)^y + (–1)^x
[#permalink]
17 Mar 2019, 05:57






