If p is a positive odd number.....
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Statement 1: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.
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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GIven: P is a positive odd integer.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.
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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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.
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.