I'm not 100% sure on this but here is my thinking:

[spoiler]D because a and b is always positive therefore it isn't the least vauluable. C can happen a lot because if you pick 0 once and the answer is zero. you can't get a -(10)^20 so that's how I came to my answer[/spoiler]

I like to see other's approach and see if I was right.[/spoiler]

I think it is (E) -(10)^20

Theory: (Negative number) ^ to (odd number) = negative & (Negative number) ^ to (even number) = postive


Logic:

As soon as I read the question, I know that a negative number raised to an odd power is negative. The biggest negative number is -10

If I multiply it 10 times I will get a positive number; but if I multiply it 9 times I get -(10)^9 which is negative. To keep the negative sign, the last number must be positive and the greatest positive number is 10.

-(10) ^ 9 * 10 = -(10)^10.

U cant factor out the - like u r saying it has to be -10^19 x 1.

What u r saying is -1 times -10^9 times 10 which would b positive

seal4913,

What does (-10)^19 * 10 equal? Certainly, not 10^20.


I am not doing any sort of factoring. These numbers cannot be factored because the base is different.

Perhaps if I do an extra step it may make more sense.

(-10)^19 * 10^(1) (that is a total of 20 numbers)
= (-1) * 10^19 * 10 ^(1)= (-1) * 10 ^20 = -(10)^20th

in the (-1) * 10^19th step that is in bold, I expanded the (-10) ^19 term in that format so it has the same base as 10^(1) so I can factor it.


Hope that explains it better

I'd go with E. To find the lowest possible product, we know it has to be negative.

The largest possible product we can come up with is 10^20 (or -10^20). How do we make it negative? Well, what if we had 19 10's and 1 -10? Like this:

10^19 * -10

We can actually break apart the -10:

10^19 * 10 * -1

And then recombine:

10^20 * -1

Which gives us:

-(10)^20