426 -> 4 + 2 + 6 = 12 12 (mod 3) = 0 (12 / 3 = 4) 426 (mod 3) = 0 (426 / 3 = 142)

This fact is actually quite simple to prove. Consider the breakdown of 426 to multiples of powers of 10:

426 = 4 * 100 + 2 * 10 + 6

Written another way:

This clearly shows that for 426 to be divisible by 3, (4 + 2 + 6) must be divisible for 3. If this isn't immediately obvious, recall that:426 = 4 * (99 + 1) + 2 * (9 + 1) + 6 = (4 * 99 + 2 * 9) + (4 + 2 + 6)

- If
`X (mod N) = 0`

, then`Y * X (mod N) = 0`

for any X and Y - If
`X (mod N) = 0`

and`Y (mod N) = 0`

, then`X + Y (mod N) = 0`

for any X and Y

P.S.:

- This theorem is, of course, an
*if and only*if relation. IFF the sum of the digits is divisible by 3, the number is divisible by 3. - Another rule says that if the sum of the digits is divisible by 9, the number is divisible by 9. Just replace 3 by 9 in the proof above, and this becomes obvious (the key fact is that 999..9 is always divisible by 3 and by 9).