What is special about the multiples of 9?

Answer 1

If I learned why this is true, then I must've forgotten. I did an internet search on sum of digits and came up with a bunch of depreciation and loan pay-off formulas. So I sat down to try to figure it out.

Let's take a 2-digit number N and the digits are ab, where a & b are digits 0-9. The value of N = 10*a + b

Not knowing if it's a multiple of nine, set a + b = 9, and let's see what happens.

Since a+b=9, then b=9-a, and substitute this into the first equation:

N = 10*a + 9 - a = 9*a + 9 --> 9*(a + 1). Since a is a whole number then (a+1) is also a whole number, so N is a multiple of 9. So by setting the sum of the digits equal 9, the number is constrained to be a multiple of nine. I know this isn't a formal proof, but it does show how it works.

When you get to 99, the sum is not nine but a multiple of nine. Any higher numbers, and you have 3 digits and more to work with. There should be some sort of general proof out there somewhere for any-digit numbers, but this is a start.

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I found the following on The Math Forum @ Drexel:

Sum of Digits of Multiples of NineDate: 08/12/2004 at 05:19:50

From: Saba

Subject: number theory: multiples of 9

Why is it that when you add the individual digits of any multiple of

nine until a single digit answer is reached the answer is always nine?

Is it possible to prove this?

For example, 99 => 9 + 9 = 18 => 1 + 8 = 9

Why doesn't it work with other numbers between 1-9 either?

Date: 08/12/2004 at 10:10:23

From: Doctor Luis

Subject: Re: number theory: multiples of 9

Hi Saba,

Good job finding that pattern! The reason is that the sum of the

digits of ANY multiple of 9 is also a multiple of 9. Since you keep

adding the digits (each time getting a new multiple of 9, but a

smaller multiple), eventually you'll end up with a single digit

number. Eventually you'll get to the multiple 9 itself.

Now, how do I know that the sum of the digits is always a multiple of

9? Suppose that a number N has digits a,b,c,d,...(from right to left),

N = a + 10b + 100c + 1000d + ...

= a + (b + 9b) + (c + 99c) + (d + 999d) + ...

= (a + b + c + d + ...) + (9b + 99c + 999d + ...)

= (a + b + c + d + ...) + 9*(b + 11c + 111d + ...)

N = (sum of digits of N) + 9 * (some number)

Now, look at that equation carefully. It means that

(sum of digits of N) = N - 9 * (some number)

Since N is assumed to be a multiple of 9, we can write it in terms of

another integer k, so that N = 9k

(sum of digits of N) = 9 * k - 9 * (some number)

= 9 * (k - (some number))

= 9 * (some other number)

Since we showed that the sum of the digits is 9 times some integer,

then it is also a multiple of 9 itself.

To summarize, starting from a multiple of 9, you keep adding the

digits, each time arriving to a multiple of 9. This establishes a

chain of decreasing multiples of 9, until eventually you reach 9 (from

a two-digit multiple). Does that make sense?

It doesn't work for other integers because the chain is broken. For

example, multiples of 8 such as 56 don't add up to a multiple of 8.

Answer 2

The digital roots of all multiples of 9 is always 9.

The digital root of a number is the 1-digit repeated sum of its digits.

So, for example, if the number is 123456789, then the sum of its digits is 1+2+3+4+5+6+7+8+9=45. That is not a 1-digit sum so repeat. 4+5 = 9.

Since the digita sum is 9, 123456789 is a multiple of 9.

Answer 3

Their digits add up to a multiple of 9.