Find the number of 10-digit numbers where the sum of the digits is divisible by 5.

Guest Nov 8, 2019

#5**0 **

Your question is not clear enough ! For example, can the digits be repeated such as:

6 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15

5 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15

4 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15

4 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15

3 + 3 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15

3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 15

2 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 15........etc.

If they can be repeated, then it is a matter of adding partitions of: 5(10 parts) + 10(10 parts) + 15(10 parts).........+90(10 parts), which would be a very involved undertaking.

If they cannot be repeated, that is, you can only use each digit once such as:1234567890, then that might be very easy to calculate. Since the sum of the 10 digits [0 to 9 =45, which is divisible by 5], then the arrangement of the 10 digits wouldn't matter at all, since they will always sum up to 45.

Therefore, if you started with: 1234567890, then you will have:9 * 9! (on the assumption that all 10 digits be different from each other). So, 9 * 9! =3,265,920.

And that is my attempt. Others should take a crack at it!

Guest Nov 8, 2019