Find all School-related info fast with the new School-Specific MBA Forum

It is currently 20 Oct 2014, 03:46

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If n is a 27-digit positive integer, all of whose digits are

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 20 May 2014
Posts: 40
Followers: 0

Kudos [?]: 1 [0], given: 1

If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 23 Jul 2014, 08:48
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

69% (01:35) correct 31% (01:09) wrong based on 67 sessions
If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 27

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Explanation given:
[Reveal] Spoiler:
Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.


CONFUSION :
[Reveal] Spoiler:
DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Jul 2014, 08:55, edited 1 time in total.
Edited the question.
Moderator
Moderator
User avatar
Joined: 25 Apr 2012
Posts: 665
Location: India
GPA: 3.21
WE: Business Development (Other)
Followers: 18

Kudos [?]: 316 [0], given: 670

Premium Member CAT Tests
Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 23 Jul 2014, 23:19
sagnik2422 wrote:
If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 27

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Explanation given:
[Reveal] Spoiler:
Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.


CONFUSION :
[Reveal] Spoiler:
DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?


Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3

Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times.

Rules for divisibility by 27: The sum of the digits should a multiple of 27

Consider no 11111111...27 times = The sum 27*1=27----> divisbible by 3,9 and 27

consider number to be 222....27 times, then sum of the no. 27*2=54 divisibly by 3,9 and 27

So why so because when you sum the numbers either you can add the digits 27 times or multiply the digit *27..

Note that since 27 is divisble by 27,9 and 3 and thus the sum of the nos will be divisible by all the nos.

Ans is E

More on this: Refer to Number properties of GMAT CLUB Math Book
math-number-theory-88376.html
_________________


“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Manager
Manager
avatar
Joined: 28 May 2014
Posts: 63
Followers: 0

Kudos [?]: 0 [0], given: 31

Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 09 Aug 2014, 08:36
Wounded Tiger,

Do we have divisibility rule for 27 as "The sum of the digits should a multiple of 27"? Say we have 54 whose sum of digits is 9. Is 9 divisible by 27? Can you explain in detail? Thanks
1 KUDOS received
Tutor
User avatar
Joined: 20 Apr 2012
Posts: 97
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Followers: 11

Kudos [?]: 119 [1] , given: 30

GMAT ToolKit User
Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 09 Aug 2014, 09:31
1
This post received
KUDOS
WoundedTiger wrote:
sagnik2422 wrote:
If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 27

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Explanation given:
[Reveal] Spoiler:
Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.


CONFUSION :
[Reveal] Spoiler:
DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?


Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3

Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times.

Rules for divisibility by 27: The sum of the digits should a multiple of 27

Consider no 11111111...27 times = The sum 27*1=27----> divisbible by 3,9 and 27

consider number to be 222....27 times, then sum of the no. 27*2=54 divisibly by 3,9 and 27

So why so because when you sum the numbers either you can add the digits 27 times or multiply the digit *27..

Note that since 27 is divisble by 27,9 and 3 and thus the sum of the nos will be divisible by all the nos.

Ans is E

More on this: Refer to Number properties of GMAT CLUB Math Book
math-number-theory-88376.html



Definitely not right criteria for divisibility by 27.
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

Tutor
User avatar
Joined: 20 Apr 2012
Posts: 97
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Followers: 11

Kudos [?]: 119 [0], given: 30

GMAT ToolKit User
Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 09 Aug 2014, 09:34
sagnik2422 wrote:
If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 27

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Explanation given:
[Reveal] Spoiler:
Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.


CONFUSION :
[Reveal] Spoiler:
DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?



I think the best solution is provided in spoiler.
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

Moderator
Moderator
User avatar
Joined: 25 Apr 2012
Posts: 665
Location: India
GPA: 3.21
WE: Business Development (Other)
Followers: 18

Kudos [?]: 316 [0], given: 670

Premium Member CAT Tests
Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 10 Aug 2014, 20:19
sri30kanth wrote:
Wounded Tiger,

Do we have divisibility rule for 27 as "The sum of the digits should a multiple of 27"? Say we have 54 whose sum of digits is 9. Is 9 divisible by 27? Can you explain in detail? Thanks



Hmmm....I agree with smyarga...

Look in general, we know a number is divisible by 3 if the sum of the digits of the nos is divisble by 3

For a number to divisible by 9, either the sum of the nos should be divisble by 9 or the number should be divisble by 3 two times

Similarly, for a no to be divisible by 27, it should be divisble by 3 three times...

There is no other method...

In my approach initially, I just extended the rule for 3 and 9 to 27 and as it can be seen it does not work out...Just went with flow I guess...

Like smyarga said, Best approach is the one given in the spoiler...
_________________


“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

VP
VP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1064
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 6

Kudos [?]: 296 [0], given: 170

Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 11 Aug 2014, 00:42
3 & 9 are multiples of 27; so divisibility applicable for 27, 3 & 9 would inherit from it

Lets take n = 2 (Neither divisible by 3, nor 6 nor 27)

Addition = 2*27 = 54 which is divisible by both 3,9 & 27

Answer = E
_________________

Kindly press "+1 Kudos" to appreciate :)

VP
VP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1064
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 6

Kudos [?]: 296 [0], given: 170

Re: If n is a 27-digit positive integer, all of whose digits are [#permalink] New post 11 Aug 2014, 00:46
sagnik2422 wrote:
If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 27

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Explanation given:
[Reveal] Spoiler:
Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.


CONFUSION :
[Reveal] Spoiler:
DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?


Its not necessary to write the whole number. Though I have taken n=2 in my earlier post, its not necessary as well. We can also solve it by variable

nnn...........n ......... 27 times

Addition of digits = n+n+n+ ........... 27 times = 27n

27n is divisible by 27; so obviously by 3 & 9

Answer =E
_________________

Kindly press "+1 Kudos" to appreciate :)

Re: If n is a 27-digit positive integer, all of whose digits are   [#permalink] 11 Aug 2014, 00:46
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic What is the two-digit positive integer whose tens digit is a babusona 2 08 Mar 2012, 06:30
7 Experts publish their posts in the topic If N is a positive Integer, is the units digit of N equal to meaglesp345 6 20 Mar 2010, 13:57
If N is a positive integer, is the units digit of N equal to showtime12 1 12 Nov 2007, 19:58
How many 5-digit positive integers exist the sum of whose apollo168 5 05 Aug 2006, 05:19
Experts publish their posts in the topic If n is a positive integer, what is the tens digit of n ? mbassmbass04 4 18 Dec 2004, 16:39
Display posts from previous: Sort by

If n is a 27-digit positive integer, all of whose digits are

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.