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

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

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29 Dec 2018, 01:20
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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]

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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

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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]

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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 :
m*n instead of x*y
_________________

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

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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]

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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]

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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]

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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
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