Observation about multiples of $999$

Question

I think I have observed an interesting pattern in the multiples of $999$. I have observed that the sum of digits of first $30$ multiples of $999$ is $27$. I have read this but it doesn't prove it. So, is it true that the sum of digits of all multiples of $999$ is a multiple of $27$? Please prove or disprove. What is the minimum multiple of $999$ that has a sum other than $27$?

Answer 1

Here are some ideas that might help answer the question:

What is the minimum multiple of 999 that has a sum other than 27?

For some integer $k$, $999 \cdot k = 1000k - k$. The digit sum of $999k$ should be greater than $27$ for at least one four-digit number $k$. If we take $k = 9999$, the digit sum of $1000k$ is already $36$, which is much greater than $27$ already, and in fact $9999 \cdot 999$ has a digit sum greater than $27$.

Glaringly obviously, with $k=1001$ we have $999 \ 999$ which has a sum of $54$. This proves that if there is any smaller $k$, it must be three digits or shorter.

But according to Szeto's answer, the sum of digits for any three-digit $k$ is $27$. For one-digit $k$, we can construct a similar argument: with $k = a$, $999k = \overline{a000} - a$. Then we have the following subtraction:

$$\begin{array}{r} &^1a \quad \quad \quad ^10\quad \ \ \ \ ^10 \quad \ \ \ \ ^10\\ -\!\!\!\!\!\!&a\\ \hline &a-1 \quad \quad \quad 9 \quad \quad \ \ 9 \ \ 10-a \end{array}$$

and the sum of digits is $a + 9 + 9 + (10-a) = 27$.

For two-digit $k$, we can construct a similar argument: $$\begin{array}{r} &a \quad \quad \quad ^1b \quad \quad \quad ^10\quad \ \ \ \ ^10 \quad \ \ \ \ ^10\\ -\!\!\!\!\!\!&a \quad \quad b\\ \hline &a \quad \quad \quad b-1 \quad \quad \ \ 9 \quad \quad 9-a \quad 10-b \end{array}$$

and the sum of digits is $a + (b-1) + 9 + (9-a) + (10-b)= 27$.

Therefore $k = 1001$ should be the smallest $k$.

What still needs to be done is to prove this rigorously for all $k$, instead of just by guessing and checking.

Answer 2

It's wrong. Take $$999\cdot2592=2589408.$$

Also, there is $$1021\cdot999=1019979.$$

Answer 3

Other examples to disprove that the sum of the digits of the multiples of $999$ is always $27$:

I hope it has helped you...

Answer 4

When you subtract two numbers, the sum of the digits of the result differs from the difference of the sums of the digits of the arguments by nine times the number of borrows performed during the subtraction.

E.g.

$$684-123=561\leftrightarrow 18-6=12\text{ (no borrow)}$$

$$623-168=455\leftrightarrow 11-15=14\color{green}{-2\cdot9}\text{ (two borrows)}.$$

Hence as a multiple of $999$ is $1000$ times a number minus that number, the sum of the digits is $9$ times the number of borrows. As the first argument ends in $3$ zeroes, there are at least three borrows, and this explains the $27$.

E.g.

$$34000-34=33966\leftrightarrow 7-7=27\color{green}{-3\cdot9}.$$

For every multiple of $999$ formed by a 3-digits number, there cannot be more than $3$ borrows (and not less).

The first case of more than $3$ borrows is with the second 4-digits number: $1001\cdot999=1001000-1001=999999$.

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For the same reason, the multiples of $99$, up to $100\cdot99$ have a sum of digits equal to $18$.

Answer 5

Let $\overline{x_1x_2...x_n}$ denote the number with decimal digits $x_1$, $x_2$, ..., $x_n$ in this order. Let's focus on at most three digits numbers $\overline{abc}$ (so it can happen that $a=0$). We can assume that $c >0$ because otherwise it would be just ten times a two digit number and multiplying by ten doesn't change the sum of digits. We have: $$999 \cdot \overline{abc}=1000\overline{abc}-\overline{abc}=\overline{abc000}-\overline{000abc}=\overline{ab(c-1)(10-a-1)(10-b-1)(10-c)}$$ So the sum of the digits is $a+b+(c-1)+(10-a-1)+(10-b-1)+(10-c)=27$. For four digits number we easily get that for $1000$ we get $999000$ which also has sum $27$ but for $1001$ we get $999999$ which is the first number to has a sum greater than $27$ but still divisible by it. Actually for numers of the form $\overline{100c}$ where $2 \le c \le 10$ we have: $$999 \cdot \overline{100c} = 999 \cdot 1000 + 999c=1000000+999(c-1) - 1$$ And since the sum of the digits of $999(c-1)$ is $27$ and it doesn't and with a zero and doesn't have more than four digits we can say that the above number has $27$ as a sum of its digits too. But the very next number: $1011$ is the first to stop obeying this rule: $$999 \cdot 1011 = 1009989$$ With the sum of the digits of $36$. Many other four digits number don't satisfy it as well.