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M and N are integers. Find the remainder when is divided by

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M and N are integers. Find the remainder when is divided by [#permalink] New post 27 Sep 2003, 10:58
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M and N are integers. Find the remainder when [M^2+N^2] is divided by 4.

(1) M and N are consequtive integers
(2) M is odd and N is even
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 [#permalink] New post 27 Sep 2003, 16:14
D -

a. 2N^2 + 2N + 1 : odd
b. (odd)^2 + (even^2) : odd

both can't be divisible by 4.
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 [#permalink] New post 28 Sep 2003, 06:34
The question is to find the remainder, and not to define whether the sum is divisible by 4 or not.
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 [#permalink] New post 29 Sep 2003, 02:09
I would say A

'coz the sum of the squares of any 2 consecutive nos is 1 more than a multiple of 4.Hence no matter which 2 consecutive nos we take, we will always have 1 as the remainder

Egs:
2^2+3^3=13 which when divided by 4 gives 1 as the remainder
4^4+5^5=41 which when divided by 4 gives 1 as the remainder
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 [#permalink] New post 29 Sep 2003, 02:57
D both are suffecient.
1) Sum of two consecutive integers always give 1 as remainder.
2) (2m)^2 + (2n+1)^2 when M & N are even and odd
= 4(m^2 + n^2 + n) + 1
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 [#permalink] New post 04 Oct 2003, 02:33
Vicky wrote:
D both are suffecient.
1) Sum of two consecutive integers always give 1 as remainder.
2) (2m)^2 + (2n+1)^2 when M & N are even and odd
= 4(m^2 + n^2 + n) + 1


D is correct. A small addition for (1) sum of SQUARES of two consequtive integers always gives 1 in a remainder when the sum is divided by 4. Lets prove it quickly.

If M and N are consequtive, then one is even and the other is odd
Let M=2k, and N=2k+1
M┬▓+N┬▓=4k┬▓+4k┬▓+4k+1=4(2k┬▓+k)+1
  [#permalink] 04 Oct 2003, 02:33
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M and N are integers. Find the remainder when is divided by

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