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Consecutive Integers

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Consecutive Integers

by ngbrian85 » Mon Apr 02, 2012 5:21 pm
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. what is the least possible value of the product of the 20 integers ?

(A) (-10)^20
(B) (-10)^10
(C) 0
(D) -(10)^19
(E) -(10)^20

Anyone have a solution >2 minutes for this?
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by seal4913 » Mon Apr 02, 2012 5:41 pm
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]
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by gencassotis » Mon Apr 02, 2012 6:14 pm
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.
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by seal4913 » Mon Apr 02, 2012 6:45 pm
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
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by gencassotis » Mon Apr 02, 2012 7:04 pm
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
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by Bill@VeritasPrep » Mon Apr 02, 2012 7:58 pm
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
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