How many 10 digit numbers are there, such that the sum of the digits is divisible by 5?
How many 10 digit numbers are there, such that the sum of the digits is divisible by 5?
The first digit cannot be zero, so 9 ways, the next 8 digits can be anything, so there are 10 ways for each of them.
We can then only choose two digits as last digit to satisfy the condition.
There are \(\mathbf{9\times 10^8 \times 2 = 1~ 800~ 000~ 000}\) numbers.
\(\begin{array}{|c|c|} \hline \text{the sum of the digits 1 until 9} & \text{last digit to satisfy the condition }\\ \hline \ldots 0 & 0~ \text{ or } ~5 \\ \ldots 1 & 4~ \text{ or } ~9 \\ \ldots 2 & 3~ \text{ or } ~8 \\ \ldots 3 & 2~ \text{ or } ~7 \\ \ldots 4 & 1~ \text{ or } ~6 \\ \ldots 5 & 0~ \text{ or } ~5 \\ \ldots 6 & 4~ \text{ or } ~9 \\ \ldots 7 & 3~ \text{ or } ~8 \\ \ldots 8 & 2~ \text{ or } ~7 \\ \ldots 9 & 1~ \text{ or } ~6 \\ \hline \end{array}\)
Example: 1 000 000 08(1) or 1 000 000 08(6)
4 567 123 98(0) or 4 567 123 98(5)