Sum of the Digits of Multiples of 9Date: 04/08/2001 at 20:10:07 From: Jennifer Springer Subject: The number nine Why do the digits of nine times a number add up to 9? For example: 9 x 4 = 36, 3 + 6 = 9 9 x 10 = 90, 9 + 0 = 9 And why does this not work for 9 x 11? 9 x 11 = 99, 9 + 9 = 18 Why not 9? Date: 04/08/2001 at 23:08:46 From: Doctor Peterson Subject: Re: The number nine Hi, Jennifer. First, let me point out that this observation can be extended to larger numbers in two ways: (1) As the rule for identifying a number divisible by 9. See our Divisibility Rules FAQ at: http://mathforum.org/dr.math/faq/faq.divisibility.html (2) In the method of "casting out nines" to determine whether any addition or multiplication is correct: Casting Out Nines to Check Arithmetic http://mathforum.org/library/drmath/view/55926.html In the simple case, we easily prove your rule algebraically. Suppose x is any single-digit number; then 9x = (10-1)x = 10x - x This almost looks like the expanded form of a decimal number, except that the digits would have to be x and -x, and we can't have negative digits. So we can "borrow," adding 10 to the units digit and taking 1 from the tens digit: 9x = 10x - x = 10(x-1) + (10-x) This tells us that the tens digit is x-1, and the units digit is 10-x. In your example with x = 4, these are 4-1 = 3 and 10-x = 6 (making the number 36). But what is the sum of the digits? (x-1) + (10-x) = 9 Now, why doesn't this work if x > 10? Because then x-1 and 10-x would not both be single digits, so there would be additional carries or borrows needed, which would mess up the trick a bit - but not completely. The sum of the digits will still be a MULTIPLE of 9 - and that's the trick for recognizing whether a number is divisible by 9. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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