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

It is currently 23 Jun 2018, 18:25

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Let f(n) = the number of distinct factors n has. For example, f(20) =

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

Hide Tags

Manager
Manager
avatar
B
Joined: 13 Sep 2016
Posts: 124
Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 11 Oct 2016, 06:02
4
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

69% (01:58) correct 31% (01:58) wrong based on 189 sessions

HideShow timer Statistics

Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Please assist with above problem.
Expert Post
4 KUDOS received
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 5938
Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 11 Oct 2016, 07:03
4
5
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Please assist with above problem.


Hi,

You don't have to do any calculations but require to know the property of factors..

A SQUARE has odd number of factors...

So 225 is ODD..
Now check the choices..
Only D is multiple of two squares so the number of factors will be odd*odd=odd..

Rest all choices will be EVEN

100=2^2*5^2..
Number of factors= (2+1)(2+1)=3*3..
10000=2^4*5^4..
Number of factors=(4+1)(4+1)=25..
answer would be F(100).F(10000)=9*25=225..

Ans D :-D
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

1 KUDOS received
Manager
Manager
avatar
B
Joined: 13 Sep 2016
Posts: 124
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 11 Oct 2016, 07:13
1
chetan2u wrote:
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Please assist with above problem.


Hi,

You don't have to do any calculations but require to know the property of factors..

A SQUARE has odd number of factors...

So 225 being square of 15 will have ODD number of factors..
Now check the choices..
Only D is multiple of two squares so the number of factors will be odd*odd=odd..

Rest all choices will be EVEN

Ans D :-D



Thanks for reply. I am still struggling to understand the logic behind this. Can you elaborate more on this?
Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 5938
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 11 Oct 2016, 07:22
1
1
alanforde800Maximus wrote:
chetan2u wrote:
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Please assist with above problem.


Hi,

You don't have to do any calculations but require to know the property of factors..

A SQUARE has odd number of factors...

So 225 being square of 15 will have ODD number of factors..
Now check the choices..
Only D is multiple of two squares so the number of factors will be odd*odd=odd..

Rest all choices will be EVEN

Ans D :-D



Thanks for reply. I am still struggling to understand the logic behind this. Can you elaborate more on this?


Hi..

Each number can be written as sum of two factors..
Example 10..1*10;2*5;.
But a square has one set which has same integer, so only one number comes out of that set..that is why ODD.
Example 36...1*36;2*18;3*12:4*9;6*6.... here 6*6 gives only one integer 6
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

Expert Post
Top Contributor
2 KUDOS received
CEO
CEO
User avatar
P
Joined: 12 Sep 2015
Posts: 2580
Location: Canada
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 11 Oct 2016, 08:31
2
Top Contributor
I'm assuming that the periods in your post are meant to represent multiplication.
I have replaced the periods with "x"

Please go back and edit your original post.

alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10) x f(100)
b) f(100) x f(1000)
c) f(1000) x f(10000)
d) f(100) x f(10000)
e) f(10) x f(1000)



NICE RULE
If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40

----------------------------------------

Let's test a few values:
100 = (2^2)(5^2)
So, the number of positive divisors of 100 = (2+1)(2+1) = (3)(3) = 9
So, f(100) = 9

1000 = (2^3)(5^3)
So, the number of positive divisors of 1000 = (3+1)(3+1) = (4)(4) = 16
So, f(1000) = 16

10000 = (2^4)(5^4)
So, the number of positive divisors of 10000 = (4+1)(4+1) = (5)(5) = 25
So, f(10000) = 25

Which of the following products is equal to 225?
225 = 9 x 25
= f(100) x f(10000)

Answer:



_________________

Brent Hanneson – Founder of gmatprepnow.com

Image

1 KUDOS received
Current Student
avatar
B
Joined: 26 Jan 2016
Posts: 110
Location: United States
GPA: 3.37
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 11 Oct 2016, 09:08
1
The easiest way to do this is to use the trick of breaking down everything into primes then looking at the exponents. Lets take 100 as an example

100 breaks down to 10*10 which is 2*5*2*5 this is 2²*5². The exponent trick is to break down a number into primes and then add one to each exponent, the multiply. So this would result in 3*3 or 9 factors.

If you do this for the numbers in the question you'll see 100 has 9 factors, 1000 has 16 factors, and and 100000 has 25 factors (these are all perfect squares).

So 9*25=225 hence D
1 KUDOS received
Manager
Manager
avatar
D
Joined: 17 May 2015
Posts: 232
CAT Tests
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 12 Oct 2016, 02:08
1
1
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Please assist with above problem.


Any integer can be written as a product of prime numbers.

If \(N = p^{a} q^{b} r^{c}\), then no. of factors of \(N = (a+1) \times (b+1) \times (c+1)\).

\(10 = 2^{1} \times 5^{1}\), No. of factors = (1+1)*(1+1) = 4
\(100 = 2^{2} \times 5^{2}\), No. of factors = (2+1) * (2+1) = 9
\(1000 = 2^{3} \times 5^{3}\), No. of factors = (3+1) * (3+1) = 16
\(10000 = 2^{4} \times 5^{4}\), No. of factors = (4+1) * (4+1) = 25

Option (d) is the correct answer.
Expert Post
EMPOWERgmat Instructor
User avatar
D
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11823
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 26 Jan 2018, 14:42
Hi All,

In this prompt, you can use "design" of this question, and the answer choices, to your advantage. We're told to find a product of two values that equals 225. In this prompt, there aren't that many ways to get to that product in this way though:

1 x 225
5 x 45
9 x 25
15 x 15

Looking at the answer choices, we know that each of those numbers has MORE than 1 factor, so (1x225) is out. We also know that none of those answers is the product of the same term twice, so (15x15) is out.

f(10) = 1,2,5,10 = 4 terms - which isn't an option, so we can eliminate Answers A and E. f(100) clearly has more than 5 factors, so we're looking for an answer that is (9x25).

From here, you just have to factor down two of the three numbers: 100, 1000, 10000 - you'll either have the exact math OR you'll have one of the correct terms and the remaining term will be the other one that you need. Thus, you'll know which answer is the match.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Senior Manager
Senior Manager
avatar
G
Joined: 31 Jul 2017
Posts: 368
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

Show Tags

New post 26 Jan 2018, 23:51
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Please assist with above problem.


10 - 4 Factors (So, Option A & E ruled out).
100 - 9 Factors
1000 - 16 Factors (Option C & B ruled out).


Hence,\(D.\)
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Re: Let f(n) = the number of distinct factors n has. For example, f(20) =   [#permalink] 26 Jan 2018, 23:51
Display posts from previous: Sort by

Let f(n) = the number of distinct factors n has. For example, f(20) =

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


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.