p is a positive odd integer,

ska7945 · Posted: Fri Oct 03, 2008 7:39 pm Post subject: p is a positive odd integer,

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.

raju232007 · Posted: Fri Oct 03, 2008 11:22 pm Post subject:

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..

sethids · Posted: Sat Oct 04, 2008 3:36 pm Post subject:

raju232007 · Posted: Sat Oct 04, 2008 11:17 pm Post subject:

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...

4meonly · Posted: Sun Oct 05, 2008 12:19 am Post subject:

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)?

raju232007 · Posted: Sun Oct 05, 2008 1:56 am Post subject:

yeah...even the above mentioned method works perfectly fine..and its much quicker than assuming numbers

kris610 · Posted: Sat Oct 11, 2008 4:56 pm Post subject:

4meonly · Posted: Sat Oct 11, 2008 10:42 pm Post subject:

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