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

Question

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

BBD · Answer

.

gaurav2m · Answer

Why are you adding 1 at the end?

JJBrink · Answer

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"

patrapan · Answer

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 .

firas92 · Answer

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

ScottTargetTestPrep · Answer

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

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