Did you know which has any natural number is often a multiple of 3 assuming the sum of its digits is also a multiple of 3? As an illustration 411 must be a multiple of 3 since 4+1+1=6 was a multiple of 3. Complicated neat!
This is a great little technique for figuring out whether or not some large number great multiple of 3 this is. Let's try to prove this result to become sure. The proof is simple with a brand new head around the notation.
Let many with the digits xj in order to x0 be divisible by giving 3. Then we would have to show that xj +... + x0 can also be divisible by 3. The giant number can be shown by ( xj* 10^j ) +... + ( x1* 10) + x0.
This is equal to xj +... + x0 + xj* (10^j - 1) +... + x1* ( 10 - 1). The terms to the right of the x0 are terms of the form 99999, 9999, 999, etc. These are all obviously divisible by 3. Hence the remaining sum xj +... + x0 must also be divisible by 3. For this reason we're done.
The same argument can be used in reverse to show that if the sum of digits of a high is divisible by 3, then so is the whole number. It's not at all concludes the proof.
This is pretty used for determining if extremely hundreds of thousands are divisible by 3 since adding is much easier than dividing! Similar tricks are used for other multiples but more proofs are needed for that; can you find them?
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