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Re: From the consecutive integers 10 to 10 inclusive, 20
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28 Mar 2016, 02:01
Mo2men wrote: Bunuel wrote: mbaspire 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 \((10)^{20}\) Hi Bunuel, I'm confused. I have referred to gmat club math book about multiply 2 number with exponents. the base should be the same to allow sum of the exponent while in this question the base is 10 & 10. So we can't sum 10^10 * (10)^19. Do I miss something? Thanks Notice that (10)^19 = (1)*(10)^19. This way you'll get the same base. Hope it's clear.
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Re: From the consecutive integers 10 to 10 inclusive, 20
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04 May 2016, 09:48
catty2004 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 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 is E.
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Re: From the consecutive integers 10 to 10 inclusive, 20
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04 May 2016, 10:49
catty2004 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 Just adding a bit of more information ( I am with the OA as E ) Here is something interesting for those who might have hard time with this question.. Think of the question in smaller terms Quote: From 10 to 10 ; select 4 numbers (repetition allowed) having the least possible product Possible numbers are { 10 , 9, 8, 7, 6, 5, 4, 3, 2, 1 , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Now you can select any number right ? But what is the best way to select least possible number ? Take 3 negative numbers (repetition allowed) \({10}^3\) = 1000 Can you make any smaller number than this ? Don't stress, you can't ... Now you have product of 3 negative numbers , multiplying with another negative number will result in a positive number, so that is not our objective... So select a positive number . But which number to choose 1 or 10, lets see.. \({10}^3\) x 1 => \(1000\) \({10}^3\) x 10 => \(10000\) Which is smaller 1000 or 10000 ? It's definitely 10000.... The similar logic applies to this problem.. Hope it helps someone, having difficulty understanding this problem..
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Re: From the consecutive integers 10 and 10, inclusive, 20 integers are r
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08 Dec 2016, 10:33
In order to minimize the product we can choose 19 biggest positive integers and then multiply their product by the smallest integer. That is X should be max if repetition is allowed.
We have: \(10^{19}*(10) = (1)*10^{20} = (10)^{20}\)
Answer E.



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Re: From the consecutive integers 10 and 10, inclusive, 20 integers are r
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08 Dec 2016, 10:44
(10)^20 is a positive number. So let's limit to (10)^19. now the last number should still keep this negative and minimize it further. for example, 1x (10)^19>2x(10)^19.....so on.. Therefore the minimum this can be is 10x(10)^19 = 1x(10^20)
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Re: From the consecutive integers 10 to 10 inclusive, 20
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18 Dec 2016, 07:00
i cannot even approach this kind of exercises, can someone elaborate? i would be grateful



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Re: From the consecutive integers 10 to 10 inclusive, 20
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Re: From the consecutive integers 10 to 10 inclusive, 20
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19 Dec 2016, 04:18
thank you a lot bunuel



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Re: From the consecutive integers 10 to 10 inclusive, 20
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25 Aug 2017, 10:46
Clearly answer is E See attached pic
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Re: From the consecutive integers 10 to 10 inclusive, 20
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15 Dec 2017, 00:48
catty2004 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 Here is the video solution to this interesting number properties question: https://www.veritasprep.com/gmatsoluti ... olving_208
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From the consecutive integers 10 to 10 inclusive, 20
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15 Dec 2017, 05:45
I have used a totally a different aproach that came to my mind, please tell me if my logic is correct. I got the answer right
We have number of consecutive integers from 10 to 10 INCLUSIVE. This means that there 21 integers, or 21 picks we can make. Pick 10 or 10 and rise it to power of 20, since we can pick one number for 20 time and then multiply the positive number you get by (1) which is the 21st final pick you can make, so you will get:
+10ˆ20 * (1) =  (10)ˆ20
I arrived to the correct answer. Is my approach/thinking correct?



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Re: From the consecutive integers 10 to 10 inclusive, 20
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05 Jan 2018, 08:24
Bunuel wrote: catty2004 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 Select 10 odd number of times and 10 the remaining number of times, for example select 10 once and 10 nineteen times, then the product will be \(10^1*(10)^{19}=10^{20}\). Answer: E. P.S. Please read and follow: http://gmatclub.com/forum/rulesforpos ... 33935.htmlHello Bunuel, hope you have a fantastc day Can you explain pleaseeee,what is the difference between (10)^20 and –(10)^20 ? yes I see minus sign – that’s why I don’t understand…. Also how can this number be the least number/ smallest one –(10)^20 if we multiply 10 TWENTY times by itself we get huge number... Thanks!



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Re: From the consecutive integers 10 to 10 inclusive, 20
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05 Jan 2018, 08:31
dave13 wrote: Bunuel wrote: catty2004 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 Select 10 odd number of times and 10 the remaining number of times, for example select 10 once and 10 nineteen times, then the product will be \(10^1*(10)^{19}=10^{20}\). Answer: E. P.S. Please read and follow: http://gmatclub.com/forum/rulesforpos ... 33935.htmlHello Bunuel, hope you have a fantastc day Can you explain pleaseeee,what is the difference between (10)^20 and –(10)^20 ? yes I see minus sign – that’s why I don’t understand…. Also how can this number be the least number/ smallest one –(10)^20 if we multiply 10 TWENTY times by itself we get huge number... Thanks! \(2^2 = (2*2) = 4\). \((2)^2 = (2)*(2) = 4\). \(10^{20}\) is (huge number), so it's has very small value. The same way as say, 100 is less than 10.
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Re: From the consecutive integers 10 to 10 inclusive, 20
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05 Jan 2018, 08:42
\(2^2 = (2*2) = 4\). \((2)^2 = (2)*(2) = 4\). \(10^{20}\) is (huge number), so it's has very small value. The same way as say, 100 is less than 10.[/quote] Bunuel thanks! one question \(10^{20}\) 10 is raised to even power which means that negative sign disappears, and number becomes positive during multiplication by itself no ?



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From the consecutive integers 10 to 10 inclusive, 20
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05 Jan 2018, 08:46
dave13 wrote: \(2^2 = (2*2) = 4\). \((2)^2 = (2)*(2) = 4\).\(10^{20}\) is (huge number), so it's has very small value. The same way as say, 100 is less than 10. Bunuel thanks! one question \(10^{20}\) 10 is raised to even power which means that negative sign disappears, and number becomes positive during multiplication by itself no ? Consider the examples in my post (highlighted). If it were \((10)^{20}\) you'd be right but it's \(10^{20}=(10*10*...*10*10*)\)
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Re: From the consecutive integers 10 to 10 inclusive, 20
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16 Nov 2018, 00:56
To get the smallest possible product, we need to consider the smallest number from this set: that would be \(10\). We are allowed to have duplicate numbers, but we have to be careful about choosing this number 20 times. If we did, then we will get: \((10)^{20}\) This ends up being a large, positive number. In order to get the smallest overall product, we need to think in terms of a negative solution. By multiplying a large product by a negative number, we can find that negative product that we're looking for. There are two possibilities: \(10 \times (10)^{19}\\ 10^{19} \times (10)\) The smallest possible product is \(–(10^{20})\). The final answer is .



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From the consecutive integers 10 to 10 inclusive, 20
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18 Dec 2018, 12:16
catty2004 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 Each of these randomly selected integers could be a negative number, a positive number, or zero, so their product could be negative. Their product has the least possible value if it is a negative number with the greatest possible absolute value. First, we select the number with the greatest possible absolute value for each place, except one. Then, we must be careful to select the right number for the last place because we have to consider not only its absolute value but also its sign. Although 10 and 10 have the same absolute value, we should choose 10s for the first 19 places because negative numbers always complicate things. Since \(10^{19}\) is positive, we must select the negative number with the greatest possible absolute value for the last place in the product. \(10^{19}(−10)=(−)10^{19}(10)=(−)10^{20}=−10^{20}\) \(−10^{20}\) is the same number as \(−(10)^{20}\) in the correct answer choice. Answer: E If the given set of integers consisted of only nonnegative or only nonpositive numbers, then we would need different strategies.
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From the consecutive integers 10 to 10 inclusive, 20
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30 Jun 2019, 10:01
mbaiseasy wrote: The minimum value is the biggest whole number with a negative sign.
\((10)*(10)^{19}\)
But this is not anywhere in the answer choice, let's manipulate: \((10)(10)^{20}=(1)(10)(10)^{19}=10^{20}\)
Answer: E IanStewartHow did we get 20 in the highlighted part? Could you help me to understand this one? Thanks__
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From the consecutive integers 10 to 10 inclusive, 20
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30 Jun 2019, 10:53
VeritasKarishmaThanks for your nice explanation. What if the word "inclusive" is removed from the question stem? VeritasKarishma wrote: catty2004 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 The consecutive integers from 10 to 10 are: 10, 9, 8, 7 ... 1, 0, 1, ... 9, 10 You have to select 20 integers with repetition. So you can select all 10. The product will be \(10^{20}\). or You can select all 10. The product will be \((10)^{20}\) (which is essentially same as \(10^{20}\)) or You can select ten 1s and ten 10s. The product will be \((10)^{10}\) or You can select 0 and any other 19 numbers. The product will be 0. or You can select 1 and nineteen 10s. The product will be \((10)^{19}\). But how will you get \( (10)^{20}\)? Editing here itself  I missed out the case in which you can get \( (10)^{20}\) (by multiplying odd number of 10 with rest of the 10s). Possibly because I misread the question. 'Least possible value of the product' to me implied the product that is not possible. Since all options a to d were possible, I overlooked that option (E) is possible too. Actually the question should read 'the smallest possible value of the product' to avoid such confusion. Since 10^20 is the smallest value that can be obtained (greatest magnitude but negative), the answer is (E).
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Re: From the consecutive integers 10 to 10 inclusive, 20
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30 Jun 2019, 12:22
Bunuel wrote: catty2004 wrote: I'm not quiet sure i understand, why choose only 10 and 10, there are other integers included in the set of 10 to 10. Also, isn't 10^1*10^(19) = 10^(18)?
We choose 10 and 10 because this way we get the least product (notice that 10^(20) is the smallest number among answer choices). Also it's \(10^1*(10)^{19}=10^{20}\) not \(10^1*(10)^{19}=10^{20}\). Hi BunuelIs it possible to prove the highlighted part?
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Re: From the consecutive integers 10 to 10 inclusive, 20
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