When $N = 111\dots1 \times 11\dots1$, what is the sum of all digits of $N$?

Question

When $N = 111\dots1 \times 111\dots1$, what is the sum of all digits of $N$? Note that the number of ones in both numbers is $1989$. Generalize if possible.

Greetings, I was doing the above question but could not solve it. This problem is tricky. I haven't made any progress on this question. The answer key tells that the answer is $17901$.

Still here's what I've done so far (at least tried so far):

Note that $\underbrace{111\dots111}_{1989\ 1\text{'s}} = \frac19 (10^{1989} - 1)$

We get $N = 11\dots11 × \frac19(10^{1989} − 1)$

I noticed that $$\frac19 \times 111...1 \text{(1989 1's)} = 12345689...123456789$$ with $221$ blocks of $123456789$.

If somehow, we could reduce the expression in terms of $\frac19 \times 111...1$, we would be done.

This is it.

EDIT: As pointed out by Robert Israel in the comments, $$(\frac{10^n−1}{9})^2 = \frac{10^{2n} - 2\cdot 10^ + 1}{81}.$$ Now $\frac{1}{81}$ repeats $.012345679$. So it's reasonable there should be a pattern that depends on $n$ mod $9$ which was pointed out earlier in this post.

Any help would be appreciated. Please try to provide the solution in layman's terms and in the easiest way.

Very Thank You.

Answer 1

From this question, we know that the sum of digits of the square of $S_n:=\sum_{k=0}^{n-1}10^k$ is equal to $81\cdot \left( \left\lfloor \frac{n}{9} \right\rfloor + \left\{\frac{n}{9}\right\}^2 \right)$.

Since $1989=9\cdot221$, we see that the sum of digits of your $N$ is equal to $9\cdot1989=17901$.

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Let me try to fill in a little gap in the hint $4$ of the referred answer.

We would like to show that $S_{n+9}^2=10^9\cdot S_n^2+S_9\cdot S_{2n+9}$.

Expanding the sum in $S_{n+9}^2$, we see that, (here a $\sum$ along means summation over $10^{k+\ell}$, abbreviated).

$$\eqalign{S_{n+9}^2&=\sum_{k=0}^{n+8}\sum_{\ell=0}^{n+8}\\ &=\sum_{k=9}^{n+8}\sum_{\ell=0}^{n-1}+\sum_{k=0}^{8}\sum_{\ell=0}^{n-1}+\sum_{k=0}^{n+8}\sum_{\ell=n}^{n+8}\\ &=10^9\cdot S_n^2+\sum_{k=0}^{8}\sum_{\ell=0}^{n-1}+\sum_{k=0}^{n-1}\sum_{\ell=n}^{n+8}+\sum_{k=n}^{n+8}\sum_{\ell=n}^{n+8}\\ &=10^9\cdot S_n^2+\sum_{k=0}^{8}\sum_{\ell=0}^{n-1}+\sum_{k=0}^{8}\sum_{\ell=n}^{2n-1}+\sum_{k=0}^{8}\sum_{\ell=2n}^{2n+8}\\ &=10^9\cdot S_n^2+\sum_{k=0}^8\sum_{\ell=0}^{2n+8}\\ &=10^9\cdot S_n^2+S_9\cdot S_{2n+9}. }$$

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Hope this helps.

Answer 2

The hint:

It's $$220\cdot37+(45+35)+219\cdot44+45=17901.$$

Number of digits of $N^2$ it's $$\left[2\log_{10}\frac{10^{1989}-1}{9}\right]+1=3977.$$

Now, our number $N^2$ it's $$AA...ABCC...CD,$$ where a number of blocks $A$ is equal to number of blocks $C$ plus $1$ and we say about following blocks: $$A=123456790,$$ $$C=987654320$$ and $$D=987654321.$$

We see that any block as $A$, $C$ and $D$ has $9$ digits.

The block $B$ it's or $0$ or $120$ or $12320$ or $1234320$ ... has odd number of digits and since $$3977=440\cdot9+17,$$ we obtain: $$B=12345678987654320.$$ Now, a sum of digits of $A$, $B$, $C$ and $D$ is equal to $37$ ,$80$, $44$ and $45$ respectively and we obtain for the needed sum of digits: $$220\cdot37+80+219\cdot44+45=17901.$$

Answer 3

Look at the multiplication table:

At the rightmost, $A=\color{red}{987654321}$ is obtained straightforward.

Next, until the borderline of $1989$ (divisible by $9$), the number of addends of $1$ is increasing and the period of $B=\color{green}{987654320}$ is repeating.

At the borderline, $C=\color{blue}{12345678}$ is the result of the change of the number of addends $1$.

After the borderline, the number of addends of $1$ starts to decrease and another period of $D=\color{purple}{123456790}$ is repeating until the leftmost.

Hence: $$A+xB+C+yD=\\ \color{red}{\frac{1+9}{2}\cdot 9}+\frac{1989-9}{9}\cdot \color{green}{\frac{2+9}{2}\cdot 8}+\color{blue}{\frac{1+8}{2}\cdot 8}+\frac{1989-9}{9}\cdot \left(\color{purple}{\frac{1+7}{2}\cdot 7+9}\right)=\\ 45+220\cdot 44+36+220\cdot 37=\\ 17,901.$$