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If p is a positive odd number.....

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Source: — Data Sufficiency |
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by GMATGuruNY » Wed Jun 01, 2016 6:49 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, the remainder is 5.

(2) p is the sum of the squares of two positive integers.
Statement 1:
In other words, p is 5 more than a multiple of 8:
p = 8a + 5 = 5, 13, 21, 29, 37...

If p=5, then p/4 = 5/4 = 1 R1.
If p=13, then p/4 = 13/4 = 3 R1.
If p=21, then p/4 = 21/4 = 5 R1.
In every case, dividing p by 4 yields a remainder of 1.
SUFFICIENT.

Statement 2:
Since p is ODD, we get:
p = (even)² + (odd)².

Let 2b = the even value in red and 2c+1 = the odd value in blue.
Then:
p = (2b)² + (2c+1)² = 4b² + 4c² + 4c + 1 = 4(b² + c² + c) + 1 = (MULTIPLE OF 4) + 1.
Since p is 1 more than a multiple of 4, dividing p by 4 will yield a remainder of 1.
SUFFICIENT.

The correct answer is D.
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by OptimusPrep » Mon Jun 06, 2016 8:09 pm
nchaswal 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.

Not able to understand how statement 2 will be effective.
GIven: P is a positive odd integer.
Required: Remainder of p/4

Let us talk about Statement 2 only
p = x^2 + y^2
We are given that p = odd
Hence p = (2k)^2 + (2n+1)^2
p = 4(k^2 + n^2) + 4n + 1
Remainder of (p/4) = 1
SUFFICIENT
Does this help?
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by Matt@VeritasPrep » Tue Jun 07, 2016 11:20 pm
A useful way is thinking of an odd integer as (2*whatever + 1). Let's call "whatever" n so that this is easier to write:

any odd = 2*n + 1

Since this odd is the sum of two other integers, it must be the sum of an even and an odd. (Even + even and odd + odd both = even, so they don't work.)

That means that

p = odd² + even²
= (2n + 1)² + (2m)²
= 4n² + 4n + 1 + 4m²
= 4*(m² + n² + n) + 1
= 4*(some junk) + 1

So the remainder is 1, and we're set.
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