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

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

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New post 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
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If m and n are integers and x*y is defined as...  [#permalink]

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New post 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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Re: If m and n are integers and x*y is defined as...  [#permalink]

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New post 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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Re: If m and n are integers and x*y is defined as...  [#permalink]

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New post 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.
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Re: If m and n are integers and x*y is defined as...  [#permalink]

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New post 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
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Re: If m and n are integers and x*y is defined as...  [#permalink]

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New post 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.
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If m and n are integers and x*y is defined as...  [#permalink]

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New post 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
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If m and n are integers and x*y is defined as... &nbs [#permalink] 02 Jan 2019, 00:56
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