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)

Here was a question that I was wondering if anyone could offer a simpler method for calculating out option III?

i suppose it's safe to assume that by 'simpler' you mean simpler than doing out all the algebra - so let's assume that you know how to do out the algebra. if you don't, then please reply accordingly.

a 'simpler' approach here is to pick three totally random numbers - say a = 3, b = 8, c = 10 - and test the statement; if the statement works with these arbitrarily chosen integers, then there's an excellent probability that it will work all the time. (notice that this is NOT A VALID METHOD OF PROOF; you may find that the statement works, but are unlucky enough to have chosen precisely the numbers that make it work by coincidence; however, that's extremely unlikely. also, notice that if the statement DOESN'T work, then you HAVE proved that it's invalid.)

try those numbers:

I. 3 + 8 - 24 = 8 + 3 - 24
-13 = -13
works

II. 3 + 0 - 0 = 3
3 = 3
works

III. first find a*b = -13 (see above), and b*c = 8 + 10 - 80 = -62.
so this says
-13 + 10 - (-130) = 3 - 62 - (-186)
127 = 127
works

looks like all three are good

Ron, but you are assuming "*" to be the operator "+", it can be any other operator such as -, X or /....

So what is a general approach for this

Here Ron is not assuming * to be +. But in the question itself it is given that a*b = a+b-ab
Sorry Ron for replying on your behalf

as pointed out by the poster above (thanks), i am making no such assumption.
all i am doing is following the exact directions provided in the question prompt, which defines the "*" operator. i am just plugging the values into the provided definition; that is all.

this is why you should really like problems involving strange made-up functions symbols; if you can FOLLOW DIRECTIONS and PLUG NUMBERS INTO THE PROVIDED DEFINITION, then you will be able to solve the problems. good stuff.

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?

You can work the algebra on this to find that all three answers are true.

A@B = B@A
Expand this out and you get A + B - AB = B + A - BA, an equation that is obviously true

A @ 0 = A
Expand this one and you get A + 0 - 0A = A, equally obviously true.

(A@B)@C = A@(B@C)
This one is a bit more difficult to expand:

(A+B-AB)@C = A@(B+C-BC)
(A+B-AB) + C - C(A+B-AB) = A + (B+C-BC) – A(B+C-BC)
(A+B-AB) + C - CA - CB + ABC = A + (B+C-BC) - AB - AC +ABC

You can now see that there are the same elements on both sides, so this is also true, but certainly not obvious!

I agree with the previous post that for most people plugging in some random numbers is the easiest way to get the right answer 95% of the time on these kinds of problems. For those reaching for scores of 750+, you'll want to be able to quickly recognize associative formulas like this one. See http://en.wikipedia.org/wiki/Associativity

Nice work.