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

It is currently 06 Dec 2019, 21:37

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

If p is an integer and 1365^p is a factor of 200

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

Hide Tags

Find Similar Topics 
Senior PS Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3300
Location: India
GPA: 3.12
If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 04 May 2018, 14:57
15
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

53% (01:57) correct 47% (02:16) wrong based on 191 sessions

HideShow timer Statistics

If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0

Source: Experts Global

_________________
You've got what it takes, but it will take everything you've got
Most Helpful Expert Reply
examPAL Representative
User avatar
P
Joined: 07 Dec 2017
Posts: 1155
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 04 May 2018, 15:20
1
1
4
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0

Source: Experts Global


As the question explicitly asks about factors and we see no way to apply divisibility rules, we'll start by prime factorization.
This is a Precise approach.

1365 = 1300 + 65 = 5(260 + 13) = 5*273 = 5*3*91=5*3*7*13
Now we'll look to our answers to help us out.
p=1 definitley works so (E) is eliminated.
if p = 7 were true than 200! would need to have at least 7 of each of 3,5,7,13. Since 7*13 = 91 then 200! has all multiples of 13 smaller or equal to 91 (and certainly also those for 3,5,7.)
So (C) works and (D) is eliminated.
Could p = 15? 15*13 =195 which is still smaller than 200! So the first 15 multiples of 13 are in 200! and (B) works.
(C) is eliminated.
Could p = 16? 16*13 = 195+13=208 so 200! does not have 16 different multiples of 13. However, before eliminating (A) we should look out for the square trap - since 169 = 13*13 than it has two factors of 13 and so 200! has 16 total 13s in its factorization - one for each of the 15 multiples and one more for 13^2.
(B) is eliminated.

(A) is our answer.
_________________
General Discussion
Director
Director
avatar
P
Joined: 02 Oct 2017
Posts: 710
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 21 May 2018, 07:20
1365= 3*5*7*13

We need to find value of p
So p will depend on lowest value one can take among factors of 1365

Among factors,13 must have less occurrences than any other factor in 200!

So occurrence of 13 is 16 in 200!
And hence 16 is limiting factor for 1365^p to be factor of 200!

Give kudos if it helps

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 01 Aug 2012
Posts: 1
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 21 May 2018, 14:57
1
200! has 200/5 + 200/25 + 200/125 zeroes ~= 49
1365 -> since the answer options are not close -> Approximating as (1000)^[p] -> (10)^[3p]
So 3p<49
p~<16
Target Test Prep Representative
User avatar
V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8622
Location: United States (CA)
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 24 May 2018, 10:48
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0


1365 = 5 x 273 = 5 x 3 x 91 = 5 x 3 x 7 x 13

Since 13 is the largest prime factor of 1365, we see that the maximum value of p will be based on the number of 13’s that are factors of 200!.

We see that 13 is a factor of 200!, and, in fact, every one of the multiples of 13 less than 200 is also a factor of 200!. Since 13 x 15 = 195, we see that there are 15 multiples of 13 that are factors of 200!. A But since 169 = 13^2, we see that there is an additional factor of 13, so there are 16 factors of 13 in 200! Thus, the maximum value of p is 16.

Answer: A
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Intern
avatar
B
Joined: 04 Apr 2017
Posts: 23
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 01 Jun 2018, 09:44
ScottTargetTestPrep wrote:
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0


1365 = 5 x 273 = 5 x 3 x 91 = 5 x 3 x 7 x 13

Since 13 is the largest prime factor of 1365, we see that the maximum value of p will be based on the number of 13’s that are factors of 200!.

We see that 13 is a factor of 200!, and, in fact, every one of the multiples of 13 less than 200 is also a factor of 200!. Since 13 x 15 = 195, we see that there are 15 multiples of 13 that are factors of 200!.A But since 169 = 13^2, we see that there is an additional factor of 13, so there are 16 factors of 13 in 200! Thus, the maximum value of p is 16.

Answer: A


Hi ScottTargetTestPrep

Sorry, I didn't get the highlighted part.

I'm following the below rule strictly, do you suggest a modification for it?
Attachments

rule.JPG
rule.JPG [ 17.42 KiB | Viewed 1195 times ]

Board of Directors
User avatar
V
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3564
Reviews Badge
If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 01 Jun 2018, 10:01
1
hisho wrote:
Hi ScottTargetTestPrep

Sorry, I didn't get the highlighted part.

I'm following the below rule strictly, do you suggest a modification for it?


Hey hisho ,

No, you are not missing anything. The formula you mentioned is the another way to solve this question.

What ScottTargetTestPrep meant was since we will have every factor of 13 less than 200 being a factor of 200! because 200! when expanded will include all those factors. Based on that we calculated there are 15 factors of 13 that are less than 200. All of those when divide 200! will give us an integer.

Now, we do have 169 also as one of it's factor less than 200. But this 169 is \(13^2\), that means we have two 13's in 169. So, we will add 1 to 15 and hence get the total number of 13s = 16.

Let's solve with your approach,

\(200/13\) + \(200/169\)

15 + 1 = 16.

Does that make sense?
_________________
My LinkedIn abhimahna.
My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place
Blog: Subscribe to Question of the Day Blog
GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.
Director
Director
avatar
P
Joined: 31 Jul 2017
Posts: 508
Location: Malaysia
Schools: INSEAD Jan '19
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 02 Jun 2018, 20:17
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0

Source: Experts Global


The Factors of \(1365\) are - \(3,5,7,13\). As 13 is the highest factor for 1365, we need to find numbers from 1 to 200 which are divisible 13.

So there are 16 numbers from 1 to 200 which are divisible by 13 (i.e. 13, 26,......, 169(two counts),... 195).

Hence, 16.
Target Test Prep Representative
User avatar
V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8622
Location: United States (CA)
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 03 Jun 2018, 18:48
hisho wrote:
ScottTargetTestPrep wrote:
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0


1365 = 5 x 273 = 5 x 3 x 91 = 5 x 3 x 7 x 13

Since 13 is the largest prime factor of 1365, we see that the maximum value of p will be based on the number of 13’s that are factors of 200!.

We see that 13 is a factor of 200!, and, in fact, every one of the multiples of 13 less than 200 is also a factor of 200!. Since 13 x 15 = 195, we see that there are 15 multiples of 13 that are factors of 200!.A But since 169 = 13^2, we see that there is an additional factor of 13, so there are 16 factors of 13 in 200! Thus, the maximum value of p is 16.

Answer: A


Hi ScottTargetTestPrep

Sorry, I didn't get the highlighted part.

I'm following the below rule strictly, do you suggest a modification for it?


When we said there are 15 multiples of 13 less than 200, we included every multiple of 13 from 13 to 195. Thus 169 is also included. However, the difference between 169 and all the other multiples is that 169 = 13^2, where all the other multiples of 13 are 13^1 * k.

Thus, the reason why we said that 169 has an additional factor of 13 is because 169 as two prime factors of 13 as opposed to just 1.
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Director
Director
User avatar
D
Status: Learning stage
Joined: 01 Oct 2017
Posts: 991
WE: Supply Chain Management (Energy and Utilities)
Re: If p is an integer and 1365^p is a factor of 200  [#permalink]

Show Tags

New post 03 Jun 2018, 21:02
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?

A. 16
B. 15
C. 7
D. 1
E. 0

Source: Experts Global


Let's write 1365 in it's prime factorization form, 1365=3*5*7*13

'p' will be maximum when the power of 13 in 200! is the highest.

Now let's apply the formula for finding number powers of the prime number 13 in 200!, we have

\(\frac{200}{13}\)+\(\frac{200}{13^2}\)
=15+1
=16

hence option(A) is the correct answer.
_________________
Regards,

PKN

Rise above the storm, you will find the sunshine
GMAT Club Bot
Re: If p is an integer and 1365^p is a factor of 200   [#permalink] 03 Jun 2018, 21:02
Display posts from previous: Sort by

If p is an integer and 1365^p is a factor of 200

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





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