Did you know that any natural number is a multiple of 3 if and only if the sum of its digits is also a multiple of 3? For example 411 must be a multiple of 3 since 4+1+1=6 is a multiple of 3. Pretty neat!
This is a great little tool for figuring out whether or not some large number is a multiple of 3 or not. Let's try to prove this result just to be sure. The proof is simple once you get your head around the notation.
Let a number with the digits xj to x0 be divisible by 3. Then we have to show that xj + ... + x0 is also divisible by 3. The large number can be represented by ( xj* 10^j ) + ... + ( x1* 10) + x0.
This is equal to xj + ... + x0 + xj* (10^j - 1) + ... + x1* ( 10 - 1) . The terms on the right of the x0 are terms of the form 99999, 9999, 999, etc. These are all clearly divisible by 3. Hence the remaining sum xj + ... + x0 must also be divisible by 3. So we're done.
The same argument can be used in reverse to show that if the sum of the digits of a number is divisible by 3, then so is the whole number. This concludes the proof.
This is pretty useful for determining if extremely large numbers are divisible by 3 since adding is much easier than dividing! Similar tricks can be used for other multiples but more proofs are required for that; can you find them?
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