GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 12 Dec 2019, 11:17

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Let S be the set of all positive integer divisors of 100,000. How many

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
P
Status: Manager
Joined: 02 Nov 2018
Posts: 280
Location: Bangladesh
GMAT ToolKit User
Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 18 Mar 2019, 08:02
16
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

21% (02:26) correct 79% (02:13) wrong based on 52 sessions

HideShow timer Statistics

Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 18 Aug 2017
Posts: 5479
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 26 Mar 2019, 09:39
1
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


factors of 10^5 = 2^5*5^5; total factors = 36
i am not able to solve further this question GMATinsight ; sir please advise on the solution
Intern
Intern
User avatar
B
Joined: 03 Nov 2018
Posts: 32
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 29 Mar 2019, 04:00
36 distinct factors in S we need 2 numbers for a product, so 2C36 = 630 ?
Manager
Manager
avatar
S
Joined: 31 Oct 2018
Posts: 77
Location: India
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 31 Mar 2019, 21:16
help needed.. Total 36 factors and selecting 2 from them gives a no. not in option.

Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121
Manager
Manager
avatar
S
Joined: 28 Jan 2019
Posts: 97
Location: India
GMAT 1: 700 Q49 V36
GPA: 4
WE: Manufacturing and Production (Manufacturing)
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 31 Mar 2019, 23:29
Archit3110 wrote:
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


factors of 10^5 = 2^5*5^5; total factors = 36
i am not able to solve further this question GMATinsight ; sir please advise on the solution


How do you determine the total factors to be 36? Kindly explain..
_________________
"Luck is when preparation meets opportunity!"
Manager
Manager
avatar
S
Joined: 19 Jan 2019
Posts: 110
Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 01 Apr 2019, 00:20
3
OhsostudiousMJ wrote:
Archit3110 wrote:
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


factors of 10^5 = 2^5*5^5; total factors = 36
i am not able to solve further this question GMATinsight ; sir please advise on the solution


How do you determine the total factors to be 36? Kindly explain..


By prime factorising 1,00,000

1,00,000 can be written as 10^5
Further break 10^5 into (2*5)^5
2^5 * 5^5.. when you have a number in the form of a^p * b^q * c^r ..... The total number of factors of that number will be
(p+1)(q+1).....

So (5+1)(5+1) = 36..

Posted from my mobile device
Manager
Manager
avatar
S
Joined: 28 Jan 2019
Posts: 97
Location: India
GMAT 1: 700 Q49 V36
GPA: 4
WE: Manufacturing and Production (Manufacturing)
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 01 Apr 2019, 00:28
thyagi wrote:

By prime factorising 1,00,000

1,00,000 can be written as 10^5
Further break 10^5 into (2*5)^5
2^5 * 5^5.. when you have a number in the form of a^p * b^q * c^r ..... The total number of factors of that number will be
(p+1)(q+1).....

So (5+1)(5+1) = 36..

Posted from my mobile device


Oh yes! Thank you! Feel silly now.. :hurt:
_________________
"Luck is when preparation meets opportunity!"
Manager
Manager
avatar
G
Joined: 23 Jan 2018
Posts: 221
Location: India
WE: Information Technology (Computer Software)
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 04 Apr 2019, 11:14
chetan2u, Gladiator59, Bunuel, VeritasKarishma, generis

dear experts :

Could you please help? Like others,I am also not able to solve this.

Regards,
Arup Sarkar
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9874
Location: Pune, India
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 05 Apr 2019, 02:03
2
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121



100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)

2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager
Senior Manager
User avatar
P
Joined: 03 Mar 2017
Posts: 373
Reviews Badge CAT Tests
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 06 Apr 2019, 23:53
VeritasKarishma wrote:
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121



100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)

2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)



HI VeritasKarishma

Applogies, but I am still not able to understand the solution that you posted. Please help.

chetan2u GMATinsight gmatbusters
_________________
--------------------------------------------------------------------------------------------------------------------------
All the Gods, All the Heavens, and All the Hells lie within you.
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 18 Aug 2017
Posts: 5479
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 07 Apr 2019, 01:02
VeritasKarishma ; hello ma'am..
I am not able to understand from the highlighted part onwards..
a. how did 121 numbers come up?
b. 2^10 , 5^10 are the highest power but how did you deduce to 117 numbers :?

GMATinsight ; sir if possible kindly look into the question and assist.

VeritasKarishma wrote:
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121



100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)
2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)
CEO
CEO
User avatar
D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Reviews Badge
Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post Updated on: 07 Apr 2019, 02:51
2
1
Archit3110 wrote:
VeritasKarishma ; hello ma'am..
I am not able to understand from the highlighted part onwards..
a. how did 121 numbers come up?
b. 2^10 , 5^10 are the highest power but how did you deduce to 117 numbers :?

GMATinsight ; sir if possible kindly look into the question and assist.

VeritasKarishma wrote:
Noshad wrote:
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121



100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)
2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)





Calculation of Number of Factors
For a Number \(N=a^p*b^q*c^r*...\)where a, b, c etc are distinct prime factors of n and p, q r are their respective exponents
The number of factors of \(N = (p+1)*(q+1)*(r+1)*...\) and so on


i.e. \(100,000 = 10^5 = 2^5 * 5^5\)

hence, total factors of 100000 = (5+1)*(5+1) = 36

i.e. S = {1, \(2^1\), \(2^2\), \(5^1\), \(2^3\), \(2^1*5^1\), \(2^4\), \(2^2*5\)........100000} 36 elements in S

The biggest terms of the set are \(2^5*5^5\) and \(2^4*5^5\) and \(2^5*5^4\)

i.e. the biggest product of two terms in set S will be \(2^5*5^5\)*\(2^4*5^5\) and \(2^5*5^5\)*\(2^5*5^4\)

i.e. Biggest terms of the product will be \(2^9*5^{10}\) and \(2^{10}*5^9\)

i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^0\) i.e. 11 combinations
Similarly, i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^1\) i.e. 11 combinations
Similarly, i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^2\) i.e. 11 combinations
and so on...

i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^{10}\) i.e. 11 combinations

i.e. Total combinations = 11*11 = 121

But as already mentioned the combinations that are not possible but included here are \(2^{10}*5^{10}\) and \(2^0*5^0\) and \(2^{10}\) and \(5^{10}\)

Total Acceptable combinations = 121 - 4 = 117

Answer: Option D

Archit3110 warrior1991
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Originally posted by GMATinsight on 07 Apr 2019, 02:06.
Last edited by GMATinsight on 07 Apr 2019, 02:51, edited 2 times in total.
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 18 Aug 2017
Posts: 5479
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

Show Tags

New post 07 Apr 2019, 02:38
1
GMATinsight ; thanks for the solution :) , this surely is a unique question not sure whether is GMAT type or not as it involves a lot of combining of pairs of factors solving under 120 sec would be really challenging.. :( :shocked

Noshad ; whats the source of this question? is it really from gmat source ? I actually googled this question and it comes up to be a question from American mathematic association /olympiad, is it so?
GMAT Club Bot
Let S be the set of all positive integer divisors of 100,000. How many   [#permalink] 07 Apr 2019, 02:38
Display posts from previous: Sort by

Let S be the set of all positive integer divisors of 100,000. How many

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne