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How many 10 digit numbers are there, such that the sum of the digits is divisible by 5?

Jan 18, 2020

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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)

Jan 18, 2020