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

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

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New post 08 May 2020, 08:03
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How many seven-digit positive integers are there,which have only 2's a  [#permalink]

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New post Updated on: 11 Jun 2020, 05:54
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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.
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How many seven-digit positive integers are there,which have only 2's a  [#permalink]

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New post 01 Jun 2020, 05:14
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?
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Re: How many seven-digit positive integers are there,which have only 2's a  [#permalink]

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New post 02 Jun 2020, 09:54
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"
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Re: How many seven-digit positive integers are there,which have only 2's a  [#permalink]

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New post 02 Jun 2020, 10:24
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 .
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How many seven-digit positive integers are there,which have only 2's a  [#permalink]

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New post 02 Jun 2020, 11:15
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
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Re: How many seven-digit positive integers are there,which have only 2's a  [#permalink]

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New post 06 Jun 2020, 05:30
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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.

Answer: B
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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

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