How many positive multiples of 36 less than 100,000 can be : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 12:20

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# How many positive multiples of 36 less than 100,000 can be

Author Message
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1264
Followers: 29

Kudos [?]: 298 [0], given: 0

How many positive multiples of 36 less than 100,000 can be [#permalink]

### Show Tags

11 Aug 2006, 09:28
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many positive multiples of 36 less than 100,000 can be formed by using only 6â€™s and 9â€™s?
Director
Joined: 17 Jul 2006
Posts: 714
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

11 Aug 2006, 10:09
768.
If it's right, then I will explain later.
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 24

Kudos [?]: 273 [0], given: 0

### Show Tags

11 Aug 2006, 11:17
Is it 6 ?

For a number to be multiple of 36 number must be divisible by 4 as well as 9.

So last digit will be 6 only.
Second last digit can not be 6 because 66 is not divisible by 4. So second last digit will be 9

1 digit numbers (None) = 0
2 digit numbers (96) = 1
3 digit numbers = 1
x96, x can be 9 only because sum of digits of 696 is not divisible by 9.
4 digit numbers (None) = 1
xx96 - Four possible values of xx - 66,69,96,99. Only 6696 has sum of digits divisible by 9.
5 digit numbers (None) = 3
xxx96 - Eight possible values for xxx
66696 - Sum of digits = 33
66996 - Sum of digits = 36 - YES
69696 - Sum of digits = 36 - YES
69996 - Sum of digits = 39
96696 - Sum of digits = 36 - YES
96996 - Sum of digits = 39
99696 - Sum of digits = 39
99996 - Sum of digits = 42

Total = 1+1+1+3 = 6
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

11 Aug 2006, 12:39
there are altogather 2^4 numbers that are even because only 16 numbers with last digit 6 that are possibly divisible by 36. let me borrow dahiya's point that the last two digits must be 96. then the total possibilities are down to 2^3 = 8.

Among them i got only 6,696; 66,996; 69,696; and 96,696 divisible by 36. but i do not know the correct approach however i tried for coupple of minuets. i am curious to know.
Senior Manager
Joined: 07 Jul 2005
Posts: 404
Location: Sunnyvale, CA
Followers: 2

Kudos [?]: 13 [0], given: 0

### Show Tags

11 Aug 2006, 12:49
Combining dahiya's and professor's approach .

Dahiya, there are no 2 digit and 3 digit nos. divisble by 36.

96 (sum = 15) not divisible by 9
696 (sum = 21)
996 (sum = 24)

Rest of the 4 no.s are what professor also found.
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

11 Aug 2006, 12:53
Professor wrote:
there are altogather 2^4 numbers that are even because only 16 numbers with last digit 6 that are possibly divisible by 36. let me borrow dahiya's point that the last two digits must be 96. then the total possibilities are down to 2^3 = 8.

Among them i got only 6,696; 66,996; 69,696; and 96,696 divisible by 36. but i do not know the correct approach however i tried for coupple of minuets. i am curious to know.

i should edit my post. will do that later. i had to run therefore my post was incomplet.

i am curious to know the elegant approach.
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 24

Kudos [?]: 273 [0], given: 0

### Show Tags

11 Aug 2006, 12:59
sgrover wrote:
Combining dahiya's and professor's approach .

Dahiya, there are no 2 digit and 3 digit nos. divisble by 36.

96 (sum = 15) not divisible by 9
696 (sum = 21)
996 (sum = 24)

Rest of the 4 no.s are what professor also found.

These silly mistakes gonna cost me a lot.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

11 Aug 2006, 14:06
* there are 4 (or 2^2) four digit numbers that may be divisible by 36. among them only 6,696 is divisible by 36.

* there are 8 (or 2^3) five digit numbers that may be divisible by 36. Among them only 66,996; 69,696; and 96,696 are divisible by 36.

so altogather 4 numbers.
11 Aug 2006, 14:06
Display posts from previous: Sort by

# How many positive multiples of 36 less than 100,000 can be

 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®.