If the operation @ is defined as a@b=a+b-ab

[Guest]

If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

[Guest]

I get C. I and II only

[RonPurewal]

you can make pretty short work of choices i and ii by plugging into the given expression.

for choice iii, one way to attack is just to plug in a bunch of random triples of numbers for a, b, and c. if the statement works for, say, two choices of random numbers in a row, then it's worth betting that it works all the time.
try a = 1, b = 3, c = -4:
(a @ b) @ c = (1) @ -4 = 1
a @ (b @ c) = 1 @ (11) = 1
works

try a = 3, b = -5, c = -2:
(a @ b) @ c = (13) @ -2 = 37
a @ (b @ c) = 3 @ (-17) = 37
works

good enough for me

--

alternatively, you could plug directly into the given definition.
(a @ b) @ c = (a+b-ab) + c - (a+b-ab)c = a + b + c - ab - ac - bc + abc
a @ (b @ c) = a + (b+c-bc) - a(b+c-bc) = a + b + c - ab - ac - bc + abc
works
watch your signs when you use the distributive property!

--

as for choosing which approach is better for you, that depends only upon your relative skill at algebraic manipulation vs. rapid arithmetic.

[teresahaglund]

I know this is an old post. But I also ran across this problem on my GMAT practice and it has me stumped. I am determined to figure it out.
I recognize I as the commutative property.
II as the Identity Property and
III as the associative property.

I also recognize a@b=a+b-ab as the general addition rule of probability. That is the only rule that I know of that will make this statement true.

Is it that simple? Just recognizing this as the general addition rule of probability and then knowing that I, II, and III would all be true using the commutative, identity and associative properties?

I sure hope someone answers me.

[tim]

you are using WAY too much specialized (and not necessarily relevant) information to tackle this problem. just use the technique Ron suggested..