Lesson PROOF of divisibility by 3 rule
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There is a simple rule how to determine if a number is divisible by three. <BLOCKQUOTE>Add up the digits and see if the sum of those digits (a much smaller number) is divisible by three.</BLOCKQUOTE> This rule simplifies finding out if a big number is divisible by three, by reducing it to a small number. Why is this proposition true? Consider, for simplicity, a 3 digit number 'abc", such as 321. Suppose that a+b+c is divisible by 3. Our number, written as 'abc' is actually 100a+10b+c. For instance, 321 = 100*3+10*2+1. Let's rewrite 100a+10b+c as 99a+a + 9b+b + c. 99a+a + 9b+b + c = (99a + 9b) + (a + b + c) The first part is always divisible by 3 since numbers with all nines are always divisible by 3 (9 = 3*3, 99=33*3, 999=333*3 etc). That the second part (a+b+c) is divisible by 3, is a given. So, we have a sum of two parts, both of which are divisible by 3. This sum is, therefore, also divisible by three. Theorem proven.
