A number is divisible by 3 if its digits add up to 3 or 9. In this case if you juggle that number in any form, the resulting number will be divisible by 3 as well. Is this a special case only for 3? Because a multiple of three can have any digit from 0 to 9?
So,if you have a 10 digit number divisible by 3, then you can get 10! numbers divisible by 3.
In fact, if you write all the numbers from 0 to 9 in any form, the resulting number is divisible by 3 since 0+1+2+...+9 =45, which adds up to 9!!
Actually you can leave the 0 in the above case and get another set of 9! numbers which are divisible by 3..
Is there more interesting properties on this topic? Please share!!
