Divisibility by Three: ProofDate: 07/22/2003 at 01:08:12 From: Emily Cheshire Subject: Sum of the digits of multiples of 3.- PROOF Why is the sum of the digits of a multiple of 3 divisible by 3? Date: 07/22/2003 at 20:46:40 From: Doctor Barrus Subject: Re: Sum of the digits of multiples of 3.- PROOF Hey, Emily! I'm going to start with an example, and then I'll give a proof. First the example: 3 * 98 = 294, so 294 is divisible by 3. Now also 2 + 9 + 4 = 15, and 15 is also divisible by 3. Why is this? Well, looking at what each digit means in the place it's in, we see that 294 = 2*100 + 9*10 + 4*1 = 2*(99+1) + 9*(9+1) + 4*(0+1) = (2*99 + 9*9 + 4*0) + (2*1 + 9*1 + 4*1) (((*))) 294 - (2*99 + 9*9 + 4*0) = 2 + 9 + 4 So the sum of the digits, which is found on the right, is equal to 294 (which is divisible by 3) minus (2*99 + 9*9 + 4*0). Now note that 99, 9, and 0 are all divisible by 3, so their sum is, too: 2*99 + 9*9 + 4*0 = 3 * (2*33 + 9*3 + 4*0) (see how I factored a 3 out of everything?) So now the left-hand side of the (((*))) equation is one number divisible by 3 minus another number divisible by three. The answer has to be divisible by 3 too. Then the sum of the digits (the right- hand side) is equal to a number divisible by 3. PROOF: Let abc be a 3-digit number with a as the hundreds digit, b as the tens digit, and c as the units digit. (So we're not multiplying a*b*c - abc are the actual digits of the number.) Suppose that abc is divisible by 3. Then 3 is a factor of 100*a + 10*b + c, which is the same as (99+1)*a + (9+1)*b + (0+1)*c, or (99*a + 9*b + 0*c) + (1*a + 1*b + 1*c) So now abc - (99*a + 9*b + 0*c) = a + b + c And every number on the left-hand side is divisible by 3, so the sum a + b + c on the right-hand side must be divisible by 3 also. So every number that's divisible by 3 has a digit sum also divisible by 3, and by working backwards (and changing things slightly) you can show that vice versa is true, as well. This proof was for 3-digit numbers, but you can do it for any number of digits, using these same steps. I hope this helps! If you have more questions, feel free to ask them. Good luck! - Doctor Barrus, The Math Forum http://mathforum.org/dr.math/ |
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