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# If k and t are integer and k^2 - t^2 is an odd integer,

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If k and t are integer and k^2 - t^2 is an odd integer, [#permalink]

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21 Apr 2008, 02:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?

i)k+t+2
ii)k^2=2kt+t^2
iii)k^2+t^2

1)None
2)i only
3)ii only
4)iii only
5)i,ii and iii
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21 Apr 2008, 05:17
patrannn wrote:
If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?

i)k+t+2
ii)k^2=2kt+t^2
iii)k^2+t^2

1)None
2)i only
3)ii only
4)iii only
5)i,ii and iii

Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.

(i) k+t is odd => (k+t) + 2 => odd

(ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH

(iii) even square + odd square (or vice versa) = even

So Answer is 4 (iii) only
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21 Apr 2008, 06:32
alpha_plus_gamma wrote:
patrannn wrote:
If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?

i)k+t+2
ii)k^2=2kt+t^2
iii)k^2+t^2

1)None
2)i only
3)ii only
4)iii only
5)i,ii and iii

Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.

(i) k+t is odd => (k+t) + 2 => odd

(ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH

(iii) even square + odd square (or vice versa) = even

So Answer is 4 (iii) only

Even + odd is odd.

I think there is some problem with the option ii as odd can't be equal to even.
i think 'none' is the answer.
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21 Apr 2008, 14:45
None. Plug in k = 3, t = 2. 3^2 - 2^2 = 5 = odd. None of the options in the list will come up as even.
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Director
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21 Apr 2008, 19:32
vshaunak@gmail.com wrote:
alpha_plus_gamma wrote:
patrannn wrote:
If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?

i)k+t+2
ii)k^2=2kt+t^2
iii)k^2+t^2

1)None
2)i only
3)ii only
4)iii only
5)i,ii and iii

Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.

(i) k+t is odd => (k+t) + 2 => odd

(ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH

(iii) even square + odd square (or vice versa) = even

So Answer is 4 (iii) only

Even + odd is odd.

I think there is some problem with the option ii as odd can't be equal to even.
i think 'none' is the answer.

Thanks for pointing that out..I don't know when will I avoid such mistakes [:(]
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21 Apr 2008, 20:48
patrannn wrote:
If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?

i)k+t+2
ii)k^2=2kt+t^2
iii)k^2+t^2

1)None
2)i only
3)ii only
4)iii only
5)i,ii and iii

We know that k^2 - t^2 = odd = (k+t)*(k-t)
We also know that only
odd*odd = odd

This means:
(k+t) = odd & (k-t) = odd
Now, Eliminate "i" because odd + 2 will always be odd.

In "ii", I think this is what you mean:
k^2 - 2kt + t^2 = (k-t)*(k-t) = odd*odd = odd
Eliminate "ii"

In "iii", we need to work out the solution further...
Knowing that
(k+t) = odd
(k-t) = odd
There are two cases here:
odd + even = odd
even + odd = odd
This means k and t will never be odd&odd or even&even pair. They will always be either even&odd or odd&even. In other word, they will always be different.

This means
k^2 + t^2 = even^2 + odd^2 = odd
Eliminate "iii"
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21 Apr 2008, 23:29
1- None

Picked by plugging numbers 3,2
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Re: number properties   [#permalink] 21 Apr 2008, 23:29
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