# How many five-digit numbers are divisible by $5$, have equal first and last digits, and have a digit-sum divisible by $5$?

How many numbers of five digits are there to follow the following conditions?

i) The numbers are divisible by $$5$$
ii) The first and last digit of the number are same
iii) the sum of the digits of the number is divisible by $$5$$

As the first condition, the last digit of the number must be 0 or 5. Then the second condition says the first and last digit of the number must be same, so the last digit can't be 0 (as it make the first digit 0 too).

Now, as per third condition, the sum of the digits of the number should be divisible by 5. As The last and first digits are 5, the middle 3 digits should be divided by 5. So, the sum of middle digits should be 0, 5, 10 or 15.

The permutations for 0, 5 and 10 are easy to compute by "stars and bars" formula. But for 15, some of the permutations have digits more than 9, like 15 = 0 + 0 + 15 = 1 + 1 + 15 = 1 + 0 + 5 and so on.

As I said before, the number of permutation for 0 is 1. the number of permutation for 5 is (5+3-1)C(5) = 21. the number of permutation for 10 is (10+3-1)C(10) -3C2 = 66 - 3 = 63, as 10 = 0+0+10 is also a permutation.

I could not figure how to permute 15.

A number is divisible by $$5$$ if and only if the first digit is either zero or five.
We conclude that both the first digit and the last digit are $$5$$. Now, we left with three digits in the middle (starting from $$000$$ to $$999$$) which means that we have $$1000$$ choices.
From these $$1000$$ numbers, we need to exclude those whose sum of digits are not divisible by $$5$$.
The number of $$3$$-digit numbers whose sum of digits are divisible by $$5$$ is $$[{\frac{1000}{5}}]=200$$ (This is the required result).