if two numbers x and y are chosen at random from the set of first 30 natural numbers. the chance that x^2-y^2 is divisible by 3 is

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if two numbers x and y are chosen at random from the set of first 30 natural numbers. the chance that x^2-y^2 is divisible by 3 is

Mathematics
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Cha-Ching, awards in the bank.
x^2-y^2= (x+y)(x-y) so we need to check what is the chance that x+y or x-y is divisible of 3 if x is divisible of 3 (1/3 chance) than there is 1/3 chance that x+y will be divisible and in that case both are (so x-y) too if x is not divisible of 3 (2/3)than if y is not divisible of 3 (2/3) chance than either x+y or x-y will be factor of 3. If y is divisible of 3 than neither x+y and x-y will be divisible. So the answer is 1/3*1/3+2/3*2/3=5/9 I am completely not sure if this is correct :-) But I thought about it for some time.
no the answer is 47/87

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can you see where my logic failed?
"if x is divisible of 3 (1/3 chance) than there is 1/3 chance that x+y will be divisible and in that case both are (so x-y) too" why do you say 1/3 chance?
that x+3 and x-y are divisible of 3
x+3?
x+y
how does dividing x by 3 has one-third chance?? m not gettin it..
1/3*9/29+2/3*19/29= 47/87
my logic was good but I did not see that it is without replacement.
1/3 chance that x is divisible of 3 than x+y or x-y only divisible of 3 if y is divisible of 3, that is 9/29 2/3 chance that x is not divisible of 3 than if y is not divisible of 3 than either x+y or x-y will be. and that is 19/29

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