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p is a positive odd integer,

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by ska7945 » Fri Oct 03, 2008 7:39 pm
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.
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Source: — Data Sufficiency |
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by raju232007 » Fri Oct 03, 2008 11:22 pm
the first statement tells that when p is divided by 8,the remainder is 5...

so p must be 8n+5 (where n=0,1,2,3,4..etc)
p= 5,21,29,37.....etc
Now when p is divided by 4 a remainder of 1 is obtained...

therefore statement 1 is sufficient..

Statement 2 tells that p is the sum of squares of two positive integers..

p=1^2+2^2=5...5/4==remainder:1
p=3^2+5^2=34...34/4==remainder:2

But remember it is given in the question that p is a positive odd integer so the second case should not be considered....

Statement 2 is also sufficient...

Hence the ans is D...
Let me know if you still have any doubts..
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by sethids » Sat Oct 04, 2008 3:36 pm
raju232007 wrote:the first statement tells that when p is divided by 8,the remainder is 5...

so p must be 8n+5 (where n=0,1,2,3,4..etc)
p= 5,21,29,37.....etc
Now when p is divided by 4 a remainder of 1 is obtained...

therefore statement 1 is sufficient..

Statement 2 tells that p is the sum of squares of two positive integers..

p=1^2+2^2=5...5/4==remainder:1
p=3^2+5^2=34...34/4==remainder:2

But remember it is given in the question that p is a positive odd integer so the second case should not be considered....

Statement 2 is also sufficient...

Hence the ans is D...
Let me know if you still have any doubts..
I agree with your answer but its not that the second case is not to be considered. The second case you have mentioned needs some modification.
The question says that p is a positive odd number. According to statement 2 we cannot rule out p being the sum of even and odd squares.
However each such sum when divided by 4 yields a remainder of 1.
e.g. p could be 100 + 81 = 181, 4 + 9 = 13, 36 + 49 = 85 ...
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by raju232007 » Sat Oct 04, 2008 11:17 pm
You have considered the squares of consecutive integers...
9^2+10^2=181............181/4..remainder=1
2^2+3^2=13................13/4..remainder=1
6^2+7^2=85.................85/4..remainder=1

But as i have already pointed out in my previous post the condition in the question is that p should be odd...so you should never consider the the case of sum of two numbers which results in an even number...

In other words...you should never consider the sum of squares of two consecutive odd integers which results in a even number

1^2+3^2=10........10/4..remainder=2
3^2+5^2=34........34/4..remainder=2
5^2+7^2=74........74/4..remainder=2

So all the above cases should be neglected as the condition in the question states that p should be an odd integer...
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by 4meonly » Sun Oct 05, 2008 12:19 am
I agree with D

I had another approach to (1) because i hate to pick numbers

Q: p=4q+R, R=?

(1)
p=8q+5
p>0, so min value of p is 5
8q is divisible by 4 with R=0
5 is divisible by 4 with R=1
so R is always =1
SUFF

(2)
I used mentioned above approach
SUFF

D

What do you think about my approach to (1)?
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by raju232007 » Sun Oct 05, 2008 1:56 am
yeah...even the above mentioned method works perfectly fine..and its much quicker than assuming numbers
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by kris610 » Sat Oct 11, 2008 4:56 pm
4meonly wrote:I agree with D

I had another approach to (1) because i hate to pick numbers

Q: p=4q+R, R=?

(1)
p=8q+5
p>0, so min value of p is 5
8q is divisible by 4 with R=0
5 is divisible by 4 with R=1
so R is always =1
SUFF

(2)
I used mentioned above approach
SUFF

D

I agree with your approach to 1.

But for 2) picking numbers to confirm may be a better idea.

9 + 16 = 25 remainder is 1

9 + 25 = 34 remainder is 2

What do you think about my approach to (1)?
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by 4meonly » Sat Oct 11, 2008 10:42 pm
I think you arfe asking about your approach to (2) because you agree with those of (1)

p is a positive odd integer (main stem) so this e.g. is correct
9 + 16 = 25 remainder is 1

but this is not
9 + 25 = 34 remainder is 2
because 34 is even. we need odd p

HTH
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by fleurdelisse » Mon Apr 20, 2009 9:49 am
How can you more generally prove that (2) is sufficient??
You can't go and sum all possible squares that yield to something odd, what if there is some pair of squares that yield to a remainder of 3?? How do we prove that the remainer can only be 1 and not 3??
Please help.
Thanks.
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Re: p is a positive odd integer,

by Ian Stewart » Mon Apr 20, 2009 3:23 pm
ska7945 wrote:If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(2) p is the sum of the squares of two positive integers.
If you want an algebraic proof, using Statement 2 alone:

p = x^2 + y^2, where x and y are integers. Since p is odd, one of the squares must be even, one odd - let's say x is even, y is odd. Then x = 2k for some integer k, and y = 2m + 1, for some integer m. So,

p = x^2 + y^2
p = (2k)^2 + (2m + 1)^2
p = 4k^2 + 4m^2 + 4m + 1
p = 4(k^2 + m^2 + m) + 1

So p is 1 more than a multiple of 4, and the remainder will be 1 when p is divided by 4. Sufficient.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by fleurdelisse » Tue Apr 21, 2009 1:05 am
Great! Thanks a lot!
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