Multiples of $999$ have digit sum $\geq 27$

Question

How could we prove the following claim?

The sum of the digits of $k\cdot 999$ is $\ge 27$

I checked $k = 1$ up to $9$. And I found that if it's true of $d$ it's also true of $10\cdot d$.

I also checked many values with a computer, it seems to always be the case. Further we can see that the digit sum must always be a multiple of 9.

I checked how to prove 'casting out nines', but I could not apply the same method here because it's just the digit sum not the digital root. and $27$ is bigger than our base $10$.

Answer 1

Lemma. Let $n$ be an integer $\ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)\le q(n)$.

Proof. $n$ has a $k$-digit decimal expansion $n=\overline{a_ka_{k-1}\ldots a_1}$ (with $k\ge 4$ and $a_k\ge1$), then $m:=n-999\cdot 10^{k-4}$ is non-negative and has a decimal expansion $m=\overline{b_kb_{k-1}\ldots b_1}$, where $b_j=a_j$ for all $j$ except $$\begin{cases}b_k=a_k-1,b_{k-3}=a_{k-3}+1&\text{if }a_{k-3}<9\\ b_k=a_k-1,b_{k-3}=0, b_{k-2}=a_{k-2}+1&\text{if }a_{k-2}<a_{k-3}=9\\ b_k=a_k-1,b_{k-2}=b_{k-3}=0, b_{k-1}=a_{k-1}+1&\text{if }a_{k-1}<a_{k-2}=a_{k-3}=9\\ b_{k-1}=b_{k-2}=b_{k-3}=0&\text{if }a_{k-1}=a_{k-2}=a_{k-3}=9\\ \end{cases} $$ Then for the digit sum of $m$ we find accordingly $$q(m)=\begin{cases}q(n)\\q(n)-9\\q(n)-18\\q(n)-27\end{cases}\le q(n) $$ Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999\cdot 10^{k-4}$, $q(n)=27$, and we can take $m=999$. $\square$

Corollary. If $n$ is a positive multiple of $999$, then $q(n)\ge 27$.

Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $\square$

Answer 2

Just a partial answer

This is true for all 3 digit $k$.

Let $k=\overline{abc}$.

The $999k=\overline{abc000}-abc$.

When $c\ne0$:

For the difference:

Unit digit is $10-c$.

Tens digit is $9-b$.

Hundreds digit is $9-a$.

Thousands digit is $c-1$.

Ten thousands digit(?) is $b$.

Hundred thousands digit(?) is $a$.

Thus the sum of digit is exactly $27$.

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A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.

Answer 3

If you are looking for a divisibility test for $27$, take the sum of $3$-digit groups starting from the units digit and adding any needed initial zeroes to the leading group. The sum matches the original number modulo $999$, thus also congruent modulo $27$ since $27\times 37=999$. For instance

$$1{,}485{,}069 \implies 001+485+069=555=20×27+15$$

so this number fails divisibility by $27$. But since $37$ is also a factor of $999$ and $555=15\times 37$, the above number passes divisibility by $37$.

Answer 4

$9|999$ so the sum of the digits of any multiple of $999$ is a multiple of $9$. So either the sum of the digits is $9$ or $18$ or $ \ge 27$.

The sum of the digits of $999 = 27 \ge 27$.

Let $k*999$ be the very lowest positive multiple in which the sums of the digits is $\le 18$.

Bear with me:

Let $B = 999k = \sum_{i=0}^n 10^ib_i$ are suppose there are two digits $b_j$ and $b_j + 3$ so that $b_j < 9$ and $b_{j+3} > 0$.

Then $C = B - 10^i*999 = B -10^i*(1000 - 1) = \sum_{i= 0}^n 10^i c_i$ where $c_j = b_j + 1$ and and $c_{j+3} = b_{j+3} - 1$ and $c_i = b_i; i \ne j, j+3$.

So the sum of the digits of $C$ is the same of the digits of $B$ but that contradicts that $B$ is the lowest multiple of $999$ with digits adding to $18$ or less.

Now $b_n \ne 0$ so that means $b_{n-3} = 9$ and $18 > b_n + b_{n-3} \ge 10$ so none of the other digits may equal $9$. Which means if there is an non zero digit $b_j$ it must be that $j < 3$.

This also means the sum of the digits must be exactly $18$.

We don't have many possible choices for $B$. To begin with if $B$ is a multiple of $10$ then $\frac B{10}$ is a smaller multiple of $999$ with the same sums of digits. So $b_0 \ne 0$ with means either $b_3 = 0$ or $n =3$.

To spell out the options with have. $B = :$

$9009$ which isn't a multiple of $999$. or

$abc9$ where $a +b+c=9; a> 0$ (easily verified that none of the first nine multiples of $999$ are fo this form. They are all of the form $a99(9-a)$. Also $abc9 - 999 = (a-1)b(c+1)0$ and the sum is less not more.

$a0b9c$ where $a+b+c = 9; c>0; a > 0$. $a0b9c - 999= (a-1)9b9(c+1)$ so the sum of digits is 27. So $a0b9c = wv*999$ for some $wv$. We can verify know such numbers match those forms. (Probably.... It'll involve tedious case checking.)

Final option is $a009bc$ and we can probably verify no $wv*999$ or $wvz*999$ are of that form.

Theres probably a much slicker way to do this.

Answer 5

Let us consider some examples, all steps in the argumentation are then also applied on the examples:

3300652000033011
12345678987654321

(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.

In our case, we separate the groups

3.300.652.000.033.011
12.345.678.987.654.321

obtain the blocks

003 and respectively 012
300                  345
652                  678
000                  987
033                  654
011                  321

and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.

(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.

(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $\le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...

Inductively we are done.

Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $\underbrace{99\dots99}_{n\text{ digits}}$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).