GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 12 Jul 2020, 11:41 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # How many seven-digit positive integers are there,which have only 2's a

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 65194
How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

1
7 00:00

Difficulty:   65% (hard)

Question Stats: 55% (02:52) correct 45% (02:52) wrong based on 51 sessions

### HideShow timer Statistics

How many seven-digit positive integers are there,which have only 2's and/or 3's as the digits and which are multiples of 12?

A. 10
B. 11
C. 12
D. 13
E. 14

_________________
Intern  B
Joined: 19 Mar 2019
Posts: 14
How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

2
Condition 1) Our final 7 digit number has to be divisible by 12- means it has to be divisible by 3 as well as 4.
Condition 2) our final 7 digit numbers has to be made up off 2 and 3 it means, we have only 32 as last 2 digits- xxxxx32. Why? - has to be divisible by 4.

so far we have figured out that if we keep 32 as last 2 digit our final number would be divisible by 4 and now we need to find how to keep this number divisible by 3 as well. So what we can do is to sum up the digits and see whether it is divisible by 3.
once we start here we will see that the combination 2222232 which has 15(2+2+2+2+2+3+2) is the minimum sum(it cant go any lower? No think about condition 1 and 2) and similarly max sum can be 18(3+3+3+2+2+3+2). Why since 32 as last 2 digits is fixed means total sum won’t reach 21((3+3+3+3+3+3+2) even if we keep 3 for all remaining 5 digits(3*5 +5 = 20)

Now let's make cases where our final 7 digit number would be divisible by 3 and 4.
Case 1: with 15 as total sum we have only value possible- 2222232
Case 2: with 18 as sum we can have max 3 of digit 3 like 22333-32,23233-32 etc. Thus, total combination- 5!/2!3! = 10

Total combination = case1 +case2= 1 + 10 = 11

Option B
Apologies for late response guys.

Originally posted by Utkarshsinghinbox on 08 May 2020, 08:18.
Last edited by Utkarshsinghinbox on 11 Jun 2020, 05:54, edited 1 time in total.
Manager  B
Joined: 14 Apr 2018
Posts: 65
Location: India
Schools: NUS '20
How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

Utkarshsinghinbox wrote:
1) the final 7 digit number has to be divisible by 12; that means it has to be divisible by 3 and 4.
2) our final 7 digit numbers has to be made up off 2 and 3 it means, we have only 32 as last 2 digits- xxxxx32. Why? - must be divisible by 4.

Next, if we sum the 7 digits then it must be divisible by 3.
So possible combination 2222232 which is 15 and also the minimum possible sum and similarly max possible sum is 18. Why because 32 as last 2 digits are fixed and total sum won’t reach 21 value. 3*5 +5 = 20

Now with 18 as sum we can have max 3 for digit 3 which means 22333-32

Possible combination- 5!/2!3! = 10

Therefore, total possible combination = 10+1 = 11

Posted from my mobile device

Why are you adding 1 at the end?
Intern  B
Joined: 27 Apr 2020
Posts: 3
Re: How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

I believe it is to account for the "32" which can only be shown 1 way. The combination of 5!/2!3! is to account for the "22333"
Intern  Joined: 30 Aug 2011
Posts: 6
Re: How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

I believe it is to account for the "32" which can only be shown 1 way. The combination of 5!/2!3! is to account for the "22333"--- If m members of a group are identical, divide the total number of arrangements by ml So here there are 2's two and three 3's are identical .
Director  D
Joined: 16 Jan 2019
Posts: 659
Location: India
Concentration: General Management
WE: Sales (Other)
How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

gaurav2m wrote:
Utkarshsinghinbox wrote:
1) the final 7 digit number has to be divisible by 12; that means it has to be divisible by 3 and 4.
2) our final 7 digit numbers has to be made up off 2 and 3 it means, we have only 32 as last 2 digits- xxxxx32. Why? - must be divisible by 4.

Next, if we sum the 7 digits then it must be divisible by 3.
So possible combination 2222232 which is 15 and also the minimum possible sum and similarly max possible sum is 18. Why because 32 as last 2 digits are fixed and total sum won’t reach 21 value. 3*5 +5 = 20

Now with 18 as sum we can have max 3 for digit 3 which means 22333-32

Possible combination- 5!/2!3! = 10

Therefore, total possible combination = 10+1 = 11

Posted from my mobile device

Why are you adding 1 at the end?

As per the solution, there is 1 possible 7 digit multiple of 12 with the sum of digits as 15 which is 2222232

And there are 10 possible 7 digit multiples of 12 with the sum of digits as 18

So the total number of possibilities is 10+1=11

Hope its clear
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11083
Location: United States (CA)
Re: How many seven-digit positive integers are there,which have only 2's a  [#permalink]

### Show Tags

2
Bunuel wrote:
How many seven-digit positive integers are there,which have only 2's and/or 3's as the digits and which are multiples of 12?

A. 10
B. 11
C. 12
D. 13
E. 14

Solution:

In order to be a multiple of 12, it must be a multiple of 3 and also a multiple of 4. Since the divisibility rule for 4 is that the last two digits of the number form a number that is divisible by 4, we see that the last two digits of such a 7-digit number must be 32 (since 23, 22 and 33 are not divisible by 4). Furthermore, since the divisibility rule for 3 is that the sum of the digits of the number is divisible by 3, we see that the sum of the first five digits and 5 (which is the sum of the last two digits) must be a multiple of 3. Since the least sum of the first five digits is 2 x 5 = 10 and the greatest sum of the first five digits is 3 x 5 = 15, the sum of the first five digits must be 10 or 13 so that when added to 5, it’s a multiple of 3.

If the sum of the first five digits is 10, then all the digits must be 2, i.e., 22222. We see that there is only one such 7-digit number in this case, i.e., 2222232.

If the sum of the first five digits is 13, then three digits are 3 and two digits are 2, e.g., 33322. However, there are 5!/(3! x 2!) = 10 ways to arrange three 3s and two 2s. Therefore, there are 10 such 7-digit numbers in this case.

Therefore, altogether there 1 + 10 = 11 such 7-digit numbers.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

214 REVIEWS

5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE

NOW WITH GMAT VERBAL (BETA)

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews Re: How many seven-digit positive integers are there,which have only 2's a   [#permalink] 06 Jun 2020, 05:30

# How many seven-digit positive integers are there,which have only 2's a  