### 9 digit number puzzle

There is a 9 digit number. No digit are repeated and rightmost digit is divisible by 1 and right 2 digits is divisible by 2, right 3 digits is divisible by 3 and so on, finally the whole number is divisible by 9.

**Can you find out the number?**

Now, Right 4 digits are divisible by 4, so the rightmost two digits can be either 20, 40, 80 or 60.

To be divisible by 9 the digits of a number must sum to a multiple of 9. Adding the 10 digits together (0+1+2+…+8+9) gives 45 which is div by 9. However, this is a 9-digit number so we have to drop one from the ten. We can only take out 0 or 9 and still have the remaining digits sum to a multiple of 9. We’ve already shown that 0 is present so the 9 is dropped.

Right 3 digits are divisible by 3 so these 3 digits can be 120, 420, 720, 240, 540, 840, 180, 480, 780, 360.

Now to be divisible by 6 sum of the six digits should be divisible by 3 and it should also be even, which it is.

There is no rule for divisible by 7 so we just have to take digits such that they are divisible by 7, there can be many such numbers and one such number is 123567480.