Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 30 May 2008
Posts: 73

From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
08 Jul 2012, 16:29
9
This post received KUDOS
71
This post was BOOKMARKED
Question Stats:
52% (00:54) correct 48% (00:54) wrong based on 2038 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 43850

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
08 Jul 2012, 16:43
4
This post received KUDOS
Expert's post
22
This post was BOOKMARKED



Manager
Joined: 30 May 2008
Posts: 73

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
08 Jul 2012, 16:53
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: rulesforpostingpleasereadthisbeforeposting133935.htmlI'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)?



Math Expert
Joined: 02 Sep 2009
Posts: 43850

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
08 Jul 2012, 16:57
catty2004 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: rulesforpostingpleasereadthisbeforeposting133935.htmlI'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 \(10^1*(10)^{19}=10^{20}\).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 30 May 2008
Posts: 73

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
08 Jul 2012, 17:35
Thanks for pointing that out!!! Silly me. But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7944
Location: Pune, India

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
08 Jul 2012, 19:59
9
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
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).
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Expert
Joined: 02 Sep 2009
Posts: 43850

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
09 Jul 2012, 00:38
1
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
VeritasPrepKarishma 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 \( (10)^{20}\)? You need to select 1 and twenty 10s but you cannot select 21 numbers. You cannot have the product as negative \(10^{20}\). Hence (E) is not possible. E is possible if you select 10 odd number of times and 10 the remaining number of times. For example: \(10^1*(10)^{19}=10^{20}\) or \(10^3*(10)^{17}=10^{20}\) or \(10^5*(10)^{15}=10^{20}\)... or \(10^{19}*(10)^{1}=10^{20}\). So, the answer is E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7944
Location: Pune, India

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
09 Jul 2012, 01:54
catty2004 wrote: But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_< Actually, the question means "smallest possible value of the product" \(10^{20}\) is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Expert
Joined: 02 Sep 2009
Posts: 43850

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
09 Jul 2012, 01:58



Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 243
Location: India
Concentration: Technology, General Management
GPA: 3.75
WE: Sales (Telecommunications)

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
24 Jul 2012, 20:59
3
This post received KUDOS
i missed out the bracket for 10^20. I was assuming that value to be positive as a negative number raised to positive power is positive. Nice problem![/edit]
_________________
I've failed over and over and over again in my life and that is why I succeedMichael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7944
Location: Pune, India

Re: From the consecutive integers 10 to 10, inclusive.. [#permalink]
Show Tags
11 Nov 2012, 18:55
1
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
Bigred2008 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 20 integers.
A. (10)^20 B. (10)^10 C. 0 D. (10)^19 E. (10)^20 The smallest possible value of the product will be negative (since it has to be smallest) with highest absolute value i.e. the number should be as left on the number line as possible. Out of all the options, (E) is the smallest. Let's see if it is possible. If you select all 10s and only one 10, you will get \(10*(10)^{19}\) which is the same as \((10)^{20}\). So it is possible to get the product given in (E). Answer must be (E) Notice that options (A) and (B) are positive.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 13 Aug 2012
Posts: 456
Concentration: Marketing, Finance
GPA: 3.23

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
19 Dec 2012, 00:32
3
This post received KUDOS
2
This post was BOOKMARKED
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
_________________
Impossible is nothing to God.



Intern
Joined: 20 Jun 2014
Posts: 15
Location: United States
Concentration: Finance, Economics
GPA: 3.87

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
25 Aug 2014, 10:05
This question shows how important it is to carefully look at the symbols. One might be very easily deceived in thanking the question is actually saying > (10)^20.



Intern
Joined: 29 Aug 2013
Posts: 14

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
19 Nov 2014, 09:07
Hi guys,
just to make sure, picking "10" 19x and "+10" the remaining time would also work, right?
> (10)^19 * 10^1
?
Many thanks



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7944
Location: Pune, India

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
19 Nov 2014, 19:47
mott wrote: Hi guys,
just to make sure, picking "10" 19x and "+10" the remaining time would also work, right?
> (10)^19 * 10^1
?
Many thanks Yes, it does work. That is how you get \((10)^{20}\).
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 432
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
20 Jan 2015, 04:26
1
This post received KUDOS
Hi,
I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.
So what I though it that we need an odd number of negative values to have a negative value. To have the greatest negative value we need 10 (the higest possible value), the greatest number of odd times; so in out case 19. We still have to use one number, so 10 would be the final choise, as it would make the negative 19(10) 10 times greater.
However, for me it is: 10(19) * 10 (1) instead of 10^1 * 10^19 = 10^20.
I understand the 10^1, but 10(19) is not 10^19. This is where I am lost...
So, why the powers?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7944
Location: Pune, India

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
20 Jan 2015, 04:36
pacifist85 wrote: Hi,
I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.
So what I though it that we need an odd number of negative values to have a negative value. To have the greatest negative value we need 10 (the higest possible value), the greatest number of odd times; so in out case 19. We still have to use one number, so 10 would be the final choise, as it would make the negative 19(10) 10 times greater.
However, for me it is: 10(19) * 10 (1) instead of 10^1 * 10^19 = 10^20.
I understand the 10^1, but 10(19) is not 10^19. This is where I am lost...
So, why the powers? You are given that you need to find the product of all 20 numbers. When you multiply 10 by 10, you get \((10)^2\), not \(10*2\) (which is 10 + (10)) Just like you multiplied the last 10 you picked, you have to multiply all previous 10s too. So multiply 10, 19 times to get \(10*10*10...*10 = (10)^{19}\)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Expert
Joined: 02 Aug 2009
Posts: 5658

From the consecutive integers 10 to 10, inclusive, 20 integers [#permalink]
Show Tags
27 Mar 2016, 07:38
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, we have to choose all 10s and ensure the combination is such that the sign becomes ive.. \(10^{19}*(10)..\) or \((10)^{19}*10= 10^{20}\) E
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
BANGALORE/



SVP
Joined: 11 Sep 2015
Posts: 2053
Location: Canada

Re: From the consecutive integers 10 to 10, inclusive, 20 integers [#permalink]
Show Tags
27 Mar 2016, 07:38
2
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
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 \((10)^{20}\) 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 Cheers, Brent
_________________
Brent Hanneson – Founder of gmatprepnow.com



VP
Joined: 26 Mar 2013
Posts: 1430

Re: From the consecutive integers 10 to 10 inclusive, 20 [#permalink]
Show Tags
28 Mar 2016, 00:42
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




Re: From the consecutive integers 10 to 10 inclusive, 20
[#permalink]
28 Mar 2016, 00:42



Go to page
1 2
Next
[ 35 posts ]



