
Re: Can you prove this ?
Posted:
Jun 26, 1996 10:41 PM


In article <d93hfa.835814513@nada.kth.se>, HÃÂÃÂ¥kan Fahlstedt <d93hfa@nada.kth.se> wrote: >Many years ago a math teacer told me that if you add all the digits in a >number and the sum is divisible by 3 then the number is also divisible by 3. > >Can you prove this in any way and if so can you find more rules like this with >other numbers.
Well, 10^k == 1 (mod 3) for k=1,2,,,
If n = a_0 + a_1*10 + a_2*10^2 + ... a_k*10^k, where a_0, a_1,..., a_k are in {0,1...9} (ie. the units digit of n is a_0, the tens digit of n is a_1 etc)
then n == 0 (mod 3) (ie. n is divisible by 3) iff a_0 + a_1*10 + a_2*10^2 + ... + a_k*10^k == 0 (mod 3) iff a_0 + a_1 + a_2 + ... + a_k == 0 (mod 3)
Some other results.
n is divisible by 2 : if a_0 is in {0,2,4,6,8} n is divisible by 3 : if the sum of the digits is divisible by 3 n is divisible by 4 : if the last two digits are divisible by 4 n is divisible by 5 : if a_0 is 0 or 5
There are a few more.
Dan

