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# If p is a positive odd integer, what is the remainder when p

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If p is a positive odd integer, what is the remainder when p [#permalink]

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23 Jul 2007, 08:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If p is a positive odd integer, what is the remainder when p is divided by 4?

(1) When p is divided by 8, the remainder is 5.

(2) p is the sum of the squares of two positive integers.

I see how (2) is sufficient, but I thought (1) was not sufficient because let p=3. Then p/4=0 R 3. p/8=0 R 5. So does the remainder thing only apply when you get an answet that is greater than 0?
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23 Jul 2007, 10:17
briks123 wrote:
If p is a positive odd integer, what is the remainder when p is divided by 4?

(1) When p is divided by 8, the remainder is 5.

(2) p is the sum of the squares of two positive integers.

I see how (2) is sufficient, but I thought (1) was not sufficient because let p=3. Then p/4=0 R 3. p/8=0 R 5. So does the remainder thing only apply when you get an answet that is greater than 0?

Yes, when a number (a) is divided by another number (b), and the remainder is >= 0, then a is >= b. So statement 1 states that when p/8, the remainder is 5. Therefore ,we know what p is greater than 8. So p can be 13, 21, ....

Am I missing something anyone? thx.
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23 Jul 2007, 19:30
Both are sufficient.

For ur example of 3, note that reminder is 3 and it doesnt satisfy equation 1. It is not a valid choice.

consider it as 8x+4+1... 8x+4 is always divisible by 4. So the reminder is 1.

For 2, Any one can show proof?
I only plugged in some numbers to double check.
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23 Jul 2007, 19:52
St1:
p = 8q1 + 5
q1 can be 1, then p = 13. p/4 -> R = 1
q1 can be 2, then p = 21. p/4 -> R = 1
q1 can be 3, then p = 29. p/4 -> R = 1
Seems R is always 1.
Sufficient.

St2:
p = x^2 + y^2
If x = 1, y = 2, then p = 5. p/4 -> R = 1
If x = 3, y = 2, then p = 13. p/4 -> R = 1
If x = 11, y = 20, then p = 521, p/4 -> R = 1
Seems R is always 1.
Sufficient.

Ans D
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23 Jul 2007, 22:20
D

P/8 > remainder 5 gives us 13,21,29,37.....
Any of those / 4 gives a remainder of 1

Sum of x^2+y^2 where either x or y is odd will always yield an odd No. with 1 as remainder
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24 Jul 2007, 22:25
it is D

statement A gives 13,21,29....
divide by 4 gives reminder 1,1,1....

statement B gives even^2 +odd^2 to get odd number
which will always yield 1 as the remainder.
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24 Jul 2007, 23:30
ArvGMAT wrote:
it is D

statement A gives 13,21,29....
divide by 4 gives reminder 1,1,1....

statement B gives even^2 +odd^2 to get odd number
which will always yield 1 as the remainder.

Can anyone show proof without plugging in numbers for St. B???
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26 Jul 2007, 14:31
ioiio wrote:
ArvGMAT wrote:
it is D

statement A gives 13,21,29....
divide by 4 gives reminder 1,1,1....

statement B gives even^2 +odd^2 to get odd number
which will always yield 1 as the remainder.

Can anyone show proof without plugging in numbers for St. B???

I think the answer is A.

1) Is sufficient as everyone has shown. Remainder is always 1.

2) Example of Remainder = 1: 2^2 + 3^2 = 4 + 9 = 13
Example of Remainder not equal 1: 3^2 + 5^2 = 9 + 25 = 34

It doesn't say the sum of consecutive integers to allow you to assume even squared plus odd squared.
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26 Jul 2007, 16:49
emoryhopeful wrote:
ioiio wrote:
ArvGMAT wrote:
it is D

statement A gives 13,21,29....
divide by 4 gives reminder 1,1,1....

statement B gives even^2 +odd^2 to get odd number
which will always yield 1 as the remainder.

Can anyone show proof without plugging in numbers for St. B???

I think the answer is A.

1) Is sufficient as everyone has shown. Remainder is always 1.

2) Example of Remainder = 1: 2^2 + 3^2 = 4 + 9 = 13
Example of Remainder not equal 1: 3^2 + 5^2 = 9 + 25 = 34

It doesn't say the sum of consecutive integers to allow you to assume even squared plus odd squared.

But Isnt P a positive ODD integer?

34 wont fit in right? For it to be odd... it needs to be a sum of odd square and even square
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26 Jul 2007, 16:53
ioiio wrote:
ArvGMAT wrote:
it is D

statement A gives 13,21,29....
divide by 4 gives reminder 1,1,1....

statement B gives even^2 +odd^2 to get odd number
which will always yield 1 as the remainder.

Can anyone show proof without plugging in numbers for St. B???

(2n)^2 + (2m -1)^2

4n2 + 4m2 -4m +1....

dividing by 4

the reminder will always be 1 if m and n are integers.

Thats the proof. I just dont know why i dint think of it that day.
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27 Jul 2007, 09:36
ioiio wrote:
emoryhopeful wrote:
ioiio wrote:
ArvGMAT wrote:
it is D

statement A gives 13,21,29....
divide by 4 gives reminder 1,1,1....

statement B gives even^2 +odd^2 to get odd number
which will always yield 1 as the remainder.

Can anyone show proof without plugging in numbers for St. B???

I think the answer is A.

1) Is sufficient as everyone has shown. Remainder is always 1.

2) Example of Remainder = 1: 2^2 + 3^2 = 4 + 9 = 13
Example of Remainder not equal 1: 3^2 + 5^2 = 9 + 25 = 34

It doesn't say the sum of consecutive integers to allow you to assume even squared plus odd squared.

But Isnt P a positive ODD integer?

34 wont fit in right? For it to be odd... it needs to be a sum of odd square and even square

Ahh, yes, my fault. You are correct, D is the answer.
27 Jul 2007, 09:36
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