How many 6-digit numbers can be formed in which the sum if its digits is divisible by 5? > In the decimal system of numeration the number of $6$-digit numbers in which the sum of the digits is divisible by $5$ is $$(A)\space180000 \space\space(B) \space540000\space\space (C) \space 5\times10^5\space \space (D) \space \text{none}$$ First, I noted that $a+b+c+d+e+f=5,10,15\dots50$ where $a\geq1$ and $ b, c, d, e, f\geq0$. From this I got $9C5 + 14C5 +...... 54C5$, which is just too hectic too calculate and I don't know it will give the right answer or not. Then I observed that there are $900000$ total possible numbers and if I divide it by $5$, I will get the correct answer which is $180000$. Is it a coincidence or is there any logic to it? What would be the proper method to solve this?

**Hint** Suppose you have specified the first five digits of a six-digit number. How many choices are there for the sixth digit such that the sum of all six digits is divisible by $5$?