Last visit was: 25 Apr 2024, 05:54 It is currently 25 Apr 2024, 05:54

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
SORT BY:
Date
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618929 [18]
Given Kudos: 81595
Send PM
Manager
Manager
Joined: 28 Jun 2016
Posts: 153
Own Kudos [?]: 190 [2]
Given Kudos: 99
Location: Canada
Concentration: Operations, Entrepreneurship
Send PM
Manager
Manager
Joined: 01 Feb 2017
Posts: 246
Own Kudos [?]: 346 [2]
Given Kudos: 148
Send PM
Intern
Intern
Joined: 19 Jul 2018
Posts: 26
Own Kudos [?]: 4 [0]
Given Kudos: 225
Location: India
GMAT 1: 680 Q49 V33
GRE 1: Q162 V167
GPA: 3.7
Send PM
Re: How many integers between 100 and 999, inclusive, have the property th [#permalink]
acegmat123 wrote:
Permutation of its digits is a multiple of 11.

Eg: 121 is a multiple so the set includes 112,121,211

Multiples of 11 between 100 and 999:

110 is the 990 is the least and greatest multiple in the range

No. of multiples = (990-110)/11 +1 = 81

So the maximum number of integers the set can have = 81*3 = 243

Eliminate C,D,E

For multiples of 10, only two permutations are possible

Eg: 990 --> 909,990

So, subtract 9 from 243 = 234

Eliminate B


Ans is A

Posted from my mobile device


What about 132? it has 6 combinations. Why did we consider 121 to get to a maximum of 243?
Tutor
Joined: 16 Oct 2010
Posts: 14822
Own Kudos [?]: 64909 [2]
Given Kudos: 426
Location: Pune, India
Send PM
Re: How many integers between 100 and 999, inclusive, have the property th [#permalink]
2
Kudos
Expert Reply
acegmat123 wrote:
Permutation of its digits is a multiple of 11.

Eg: 121 is a multiple so the set includes 112,121,211

Multiples of 11 between 100 and 999:

110 is the 990 is the least and greatest multiple in the range

No. of multiples = (990-110)/11 +1 = 81

So the maximum number of integers the set can have = 81*3 = 243

Eliminate C,D,E

For multiples of 10, only two permutations are possible

Eg: 990 --> 909,990

So, subtract 9 from 243 = 234

Eliminate B


Ans is A

Posted from my mobile device


Note that other than numbers such as 121 (2 digits same), you will also have numbers such as 132 which are multiples of 11 (all digits distinct). You can write these in 3! ways = 6 ways. But here, one way is 231 which is another multiple of 3 so in effect, each multiple will get 3 distinct ways.
Hence your logic works in each case.
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32667
Own Kudos [?]: 821 [0]
Given Kudos: 0
Send PM
Re: How many integers between 100 and 999, inclusive, have the property th [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: How many integers between 100 and 999, inclusive, have the property th [#permalink]
Moderators:
Math Expert
92912 posts
Senior Moderator - Masters Forum
3137 posts

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