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OG 196 (Properties of numbers/exponents)

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OG 196 (Properties of numbers/exponents)

by DCS80 » Sat Jan 12, 2013 9:05 am
Hi,


#196 (OG 13)

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


Having issues grasping how D is distinguishable from E, and also how they are differentiated from A

thanks
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by GMATGuruNY » Sat Jan 12, 2013 5:02 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
Since it's possible for the product to be negative, the correct answer must be D or E.
Since the product is composed of twenty integers, the needed exponent is probably 20 -- as in answer choice E.

-10 * 10¹� = -(10²�).

The correct answer is E.
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by subhakam » Mon Jan 14, 2013 7:48 pm
Thank you - for a second i thought answer was 0 then realized they are asking for least possible value not least positive value. So how can -(10) ^20 be negative? If I multiple -10 20 times does it not come out to be positive?
Thanks
GMATGuruNY wrote:
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
Since it's possible for the product to be negative, the correct answer must be D or E.
Since the product is composed of twenty integers, the needed exponent is probably 20 -- as in answer choice E.

-10 * 10¹� = -(10²�).

The correct answer is E.
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by Jeff@TargetTestPrep » Sat Jul 18, 2015 6:56 pm
DCS80 wrote:Hi,


#196 (OG 13)

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


Having issues grasping how D is distinguishable from E, and also how they are differentiated from A

thanks
Solution:

This problem is testing our knowledge of the multiplication rules for positive and negative numbers. Remember that when we multiply an even number of negative numbers together the result is positive, and when we multiply an odd number of negative numbers together the result is negative.

Because we are selecting 20 numbers from the list, we want to start by selecting the smallest 19 negative numbers we can and multiplying those together. In our list the smallest number we can select is -10. So we have:

(-10)^19 (Note that this product will be negative.)

Because we need to select a total of 20 numbers, we must select one additional number, which must be positive, from the list. However, since the final product must be as small as possible, we want the final number we select to be the largest positive value in our list. The largest positive value in our list is 10. So the product of our 20 integers is:

(-10)^19 x 10 (Note that this product will still be negative.)

This does not look identical to any of our answer choices. However, notice that
(-10)^19 = (-1 x 10)^19 = (-1)^19 x (10)^19 = - 1 x (10)^19 = -(10)^19, so

(-10)^19 x 10 = -(10)^19 x (10)^1 = -(10)^20.

Answer: E

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by [email protected] » Sun Jul 19, 2015 6:04 am
Hi All,

This question is built around Number Properties (Number Lines) and basic arithmetic (multiplication and exponents). Since we're multiplying 20 integers together and we want the SMALLEST PRODUCT possible, we need to consider HOW we can make that small product happen.

We're given the range of -10 to +10, inclusive AND we're allowed to repeat numbers. Since the range includes negative values, we can clearly end up with a negative product - it's just a matter of how 'negative' we can make it.

Conceptually, there are two ways to get the smallest number possible (meaning the FARTHEST LEFT on the number line):

(10^19)(-10)
[(-10)^19](10)

Each of these products will have the same end value...

Final Answer: E

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by nikhilgmat31 » Wed Jul 29, 2015 4:41 am
(-10) ^19 for 19 digits & 10 for 20th digit

gives -(10)^20

Initially I selected 1 as 20th digit but my mistake. Actually 10 should 20th digit for product to be minimum.
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by Brent@GMATPrepNow » Wed Aug 23, 2017 4:42 pm
DCS80 wrote: 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
Choose nineteen -10's and one 10
So, the product = [(-10)^19][10]
Notice that [(-10)^19] is NEGATIVE, which means [(-10)^19][10] is also NEGATIVE.

So, [(-10)^19][10] = -[(10)^19][10]
= -(10)^20
= E

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by Matt@VeritasPrep » Wed Aug 30, 2017 5:51 pm
Maybe an easier way of thinking about this:

To make the number as small as possible, we want to multiply the number with the largest absolute value (in this case, either -10 or 10) times itself pretty much every time, then end with the smallest negative number (should we need it) to force the result to be negative.

In this case, that gives us nineteen 10s and one -10, or -10 * 10¹�, or E.
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by Jeff@TargetTestPrep » Mon Sep 04, 2017 10:15 am

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
This problem is testing our knowledge of the multiplication rules for positive and negative numbers. Remember that when we multiply an even number of negative numbers together, the result is positive, and when we multiply an odd number of negative numbers together, the result is negative.

Because we are selecting 20 numbers from the list, we want to start by selecting the smallest 19 numbers we can and multiplying those together. In our list, the smallest number we can select is -10. So, we have:

(-10)^19 (Note that this product will be negative.)

Since we need to select a total of 20 numbers, we must select one additional number from the list. However, since the final product must be as small as possible, we want the final number we select to be the largest positive value in our list. The largest positive value in our list is 10. So, the product of our 20 integers is:

(-10)^19 x 10 (Note that this product will still be negative.)

This does not look identical to any of our answer choices. However, notice that (-10)^19 can be rewritten as -(10)^19, so:

(-10)^19 x 10 = -(10)^19 x (10)^1 = -(10)^20

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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