Expert: Sherry Wallin - 12/2/2010

Question
15)How many five digit numbers can be formed using the digit 0,1,2,3,4 and 5 which are divisible by 3 ,without repetition of the digits

Answer
Dhananjay~

Numbers that are divisible by 3 have an interesting characteristic and that is if you add the digits of the number and that sum is divisible by 3 then the number is divisible by 3 also. Maybe we can use that in figuring out this solution.

As before, we want to calculate how many 5 digit numbers without repetition can be made from the 6 digits:

There are 5 choices for the 1st digit. Why? *
There are 5 choices for the 2nd digit. Why? **
There are 4 choices for the 3rd digit.
There are 3 choices for the 4th digit.
There are 2 choices for the 5th digit.
Therefore there are 5*5*4*3*2 = 5*5! = 600 ways to make 5 digits numbers without repetition.

Look at the 6-digits and determine what arrangements of 5-digits will give you divisibility by 3: I see that 1+2+3+4+5 = 15 which is divisible by 3, and how many different arrangements are there using these digits? Think about this.

There is another way to get a sum divisible by 3 using 5-digits:
1+2+4+5+0 = 12. How many ways can you use these 5 digits?

Since the sum of the largest numbers in the set, i.e. 1+2+3+4+5 = 15 the only sums divisible by 3 would be 0,3,6,9,12,15. Can you get a sum of 9 using 5 digits? The smallest 5 is 0+1+2+3+4 = 10 which isn't divisible by 3 so there aren't any more, just those using 1,2,3,4,5 and 0,1,2,4,5.

Now we need to determine how many of those are divisible by 3:
The number of ways using 1,2,3,4,5 is 5! and the number of ways using 0,1,2,4,5 is 5! so there are 2*5! = 240 ways to make 5 digit numbers from the digits 0,1,2,3,4,5 divisible by 3 without repetition.


Math Prof

* you can't start with 0
** you can have 0 plus the other 4 digits left after the first pick