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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
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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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Re: Another GMAT prep [#permalink]
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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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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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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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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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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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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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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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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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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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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.










