Multiples of 9 Number PuzzleDate: 03/23/2003 at 15:57:43 From: Brittney Subject: Multiple of 9 If you take a two-digit number like 21, add the two digits (2+1=3), and subtract the answer from the original number (21-3=18), why is the final answer always a multiple of 9? Date: 03/24/2003 at 03:21:21 From: Doctor Jeremiah Subject: Re: Multiple of 9 Hi Brittney, A two-digit number can be represented as multiples of 10. For example: 21 = 10x2 + 1 and 45 = 10x4 + 5. So a number with the left-hand digit of a and the right-hand digit of b can be written as 10a + b. If you sum the digits you get a + b. If you subtract that sum from the original number you get (10a + b) - (a + b) 10a - a + b - b 9a So the value is always 9 times the left-hand digit. And anything times nine is a multiple of nine. This also works with three-digit numbers. A three-digit number is 100a + 10b + c. The sum of the digits is a + b + c. If we subtract them we get: (100a + 10b + c) - (a + b + c) 100a - a + 10b - b + c - c 99a + 9b And that is also evenly divisible by nine. In fact this works for all numbers of any length. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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