# Finding how many 8 digit numbers, with a twist

• Mar 26th 2010, 11:52 AM
Monika1987
Finding how many 8 digit numbers, with a twist
I need to find how many 8 digit numbers exists that have no zeros in them and each of the other digits (1-9) may only occur once. (no repeats)

Also the numbers must be devisable with three.

Please help, i have been trying to figur this one out for a while but i am really stuck.

Regards Monika
• Mar 26th 2010, 12:47 PM
Plato
Quote:

Originally Posted by Monika1987
I need to find how many 8 digit numbers exists that have no zeros in them and each of the other digits (1-9) may only occur once. (no repeats) Also the numbers must be divisible with three.

Any number divisible three must have the sum of its digits divisible by three.
Note that $1+2+4+5+7+8=27$ thus if we add any two of $\{3,6,9\}$ we have an eight-digit number divisible by three.
Note also that those are the only eight-digit numbers divisible by three.
Rearrange each set. What is the total?