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I do not get this, Plz help!!!!

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I do not get this, Plz help!!!!

by eustudent » Wed Jun 10, 2009 5:21 am
If P is a positive odd integer, what is the remainder when P is divided by 4?

1) When P is divided by 8, reiminder is 5

2) p is the sum of the squares of two positive intgers

Answer is ____

Question:
for statement 2 we use the integers of 4^2+5^2, 3^2+ 2^2, so far so good. reminder is 1, however, let us use 6^2+8^2. we get a different reminder, hence statement 2 is insufficient
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Source: — Data Sufficiency |
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by sportcntr3 » Wed Jun 10, 2009 5:30 am
Believe the answer is A as the remainder will end up being 1, since 4 is a factor of 8

13/8 = 1 remainder 5

13/4 = 3 remainder 1

21/8 = 2 remainder 4

21/4 = 5 remainder 1

29/8 = 3 remainder 5

29/4 = 7 remainder 1

37/8 = 4 remainder 5

37/4 = 9 remainder 1



As for statement 2 I don't think that limits the options very much at all

Did that answer your question?
Last edited by sportcntr3 on Wed Jun 10, 2009 6:03 am, edited 1 time in total.
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by tohellandback » Wed Jun 10, 2009 5:56 am
IMO A
the remainder will be 1.
13/8-reminder is 5
divide by 4 and reminder will be 1.
take a few other examples
so 1 is sufficient

2^2+2^2=8 divided by 4,remainder 0
3^2+3^2=18 divided by 4 remainder 2, not sufficient
The powers of two are bloody impolite!!
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by eustudent » Wed Jun 10, 2009 6:16 am
Hi,

I also put A as an answer.

but it is D.

Just do not understand why would it be D?
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by sportcntr3 » Wed Jun 10, 2009 6:30 am
So in looking again I think its bc in order for P to both be odd and the sum of two squares it needs to be the sum of an odd square and an even square. This will then lead to the remainder again always being 1

4 + 9 = 13 (when divided by 4 has a remainder of 1)

16 + 25 = 41 (remainder 1)

36 + 81 = 117 (remainder 1)

basically bc it always need to be an even and an odd number (in order for P to be odd) the remainder will end up being 1. If someone has a more technical explanation it would probably be helpful, but it does work, so D
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by tohellandback » Wed Jun 10, 2009 6:37 am
oh boy!!
totally missed the odd part..this forum is really helping me find my demons!!!
thanks sportcntr3!

by the way now I remember this property:
any number can be represented as a sum of sqare of two numbers unless the number is of the form 4k+3
so all numbers of the form 4k+1 and 4k+2 can be represented like this
now in our case the number is odd. so it must be of the form 4k+1 and so remainder 1.

Hope I helped
The powers of two are bloody impolite!!
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by eustudent » Wed Jun 10, 2009 11:04 am
Thank you for help!

But i do not see where the problem says that they have to be odd and even square numbers.

what if i decide to use 36+64? Then, there is no reminder.

I mean how did you infer that they have to be consecutive or odd and even sqares?

Thanks
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by sportcntr3 » Wed Jun 10, 2009 11:10 am
"If P is a positive odd integer, what is the remainder when P is divided by 4?"

It says P must be a positive ODD integer. In order for that to occur, with regard to statement 2, the two positive squares must be one odd and one even as the sum of an odd number and an even number will be odd.

Odd + odd = even, even + even = even, odd + even = odd, therefore the resulting squares that sum up to P must be one odd and one even
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by eustudent » Wed Jun 10, 2009 11:25 am
I see now. Yes, I missed that part.


Thank you for your help!
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by ghacker » Wed Jun 10, 2009 12:23 pm
P is an odd integer



Statement I says when P is divided by 8 the remainder is 3 , we see that it is in the form of 4(2k)+3 hence the remainder is 3 , I is sufficient

Statement II ; P is the sum of squares of two integers ,p is also odd then the two integers must be an odd and an even integer

P = (2s)^2+(2r+1)^2 = 4(s^2+r^2+r)+1 =4K+1 , hence its also sufficient

So answer is D
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