Divisibility

[norizam]

This question is my modified version of a question found somewhere else.I hope it's ok to post it here.
When A is divided by 7, the remainder is X. When B is divided by 7, the remainder is Y. What is the value of X+Y?
(1) A + B is divisible by 7
(2) A-B is divisible by 7

my solution:
1. A + B (mod 7) ≡ X + Y ( mod 7 )
so X +Y is either 0 or 7 ∴ not sufficient

2. A - B (mod 7) ≡ X - Y ( mod 7 )
again X -Y is either 0 or 7
if X-Y =0, then X=Y ∴ X + Y could be 0+0,1+1,2+2, ... ,6+6
At this point i take it that (2) is insufficient
my question is how should I deal with x-y =7?Can x, y be numbers such that the difference is 7? If I read remainder theorem correctly, the remainder when a number is divide by 7 is between (and including)0 and 7.
My answer is C. from (2) X=y
from(1) x + y=0 or 7
∴ x + y=0

My second question is:Do remainder problems in GMAT deal only with positive numbers? What if A=10 and B=-4, then X=3 and Y=-4?
Thanks
btw-I am just about to exhaust all the how-to's from the forums. Many thanks

[michael_shaunn]

Note:The remainders on dividing A and B by 7 have been referred by the alphabets x and y repectively.Hence x not is not equal to y.(That's what i think).
Now, x lies between 0 and 6 and so does y.(Hence both are non-negative).
In order for A+B to be divisible by 7, the value of x+y must be divisible by 7.So,x+y can only be 7 and nothing else.(Why x+y is not equal to 0?)....x+y can't be 0 because both x and y are non-negative.For x+y to be zero,x and y both will have to be 0 and in that case x will be equal to y which is not the case as the remainders have been referred by different alphabets.

For A-B to be divisible by 7 the value of x-y should be divisible by 7.The value of x-y can't exceed 6 and can't be less than -6.So the only value that lies between them that is divisible by 7 is 0.Hence x is equal to y(which seems ambiguous since the remainders have been referred by differnent alphabets and should not be equal(well,that's what i think).

Somebody plz clarify.

[sharok50]

A=7p+X
B=7q+Y

St.1 Say A+B is divisible by 7
it means
A+B= 7p+X+ 7q+Y
A+B= 7(P+q)+(X+Y)
if A+B is divisible by 7 then X+Y=0

So St1 is sufficient

St2 A-B= (7p+X)- (7q+Y)

A-B= (7p-7q)+ (X-Y)
Here we know that X-Y=0

So st 2 is not sufficient

My answer is A