I think it is A.
p=8x+5=4(2x)+4+1=4(2x+1)+1
remainder when p is divided by 4 is 1.
II does not tell us anything.
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tendays2go
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it is D
for first stmt has been explained by alescau already.
for the second stmt, pick up values for the two positive integers.
Note that P is an odd integer so we have to choose the integers such that
one is odd and other is even
thus, odd^2 + even^2 = odd
let's take 1 &2 => P = 5, and remainder is 1
for 6 and 9 => P = 117 and here remainder is 1
for 4 and 13 => P = 185 thus, remainder =1
hence, both are sufficient
for first stmt has been explained by alescau already.
for the second stmt, pick up values for the two positive integers.
Note that P is an odd integer so we have to choose the integers such that
one is odd and other is even
thus, odd^2 + even^2 = odd
let's take 1 &2 => P = 5, and remainder is 1
for 6 and 9 => P = 117 and here remainder is 1
for 4 and 13 => P = 185 thus, remainder =1
hence, both are sufficient
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stubbornp
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A....
for stmt 1,5,13,21,... are the numbers
which gives same remainder 1 when divided by 4....
in case of b
3=1^2+2^2
5=2^2+1^2....
more than one value with stmt 2
thats y.....i suggest A...
for stmt 1,5,13,21,... are the numbers
which gives same remainder 1 when divided by 4....
in case of b
3=1^2+2^2
5=2^2+1^2....
more than one value with stmt 2
thats y.....i suggest A...
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4meonly
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let's take 3^2+9^2=90, R=2tendays2go wrote:
let's take 1 &2 => P = 5, and remainder is 1
for 6 and 9 => P = 117 and here remainder is 1
for 4 and 13 => P = 185 thus, remainder =1
INSUFF
As alescau posted p=8x+5=4(2x)+4+1=4(2x+1)+1
remainder when p is divided by 4 is 1.
I agree with A
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schumi_gmat
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