Repeated Concatenation

First of all, a number is divisible by 11, if and only if the alternating sum of the digits is a multiple of 11. This works because of congruence modulo 11 between every other power of 10 (see this website if you're not sure about this).

Case 1: $n$ has an odd number of digits

For example let $n = 361$ . If we concatenate it with itself once we get $\huge \color{#D61F06}{3} \color{#3D99F6}{6} \color{#D61F06}{1} \color{#3D99F6}{3} \color{#D61F06}{6} \color{#3D99F6}{1}$ $\text{Alternating Sum} = \color{#D61F06}{3} \color{#333333}{-} \color{#3D99F6}{6} \color{#333333}{+} \color{#D61F06}{1} \color{#333333}{-} \color{#3D99F6}{3} \color{#333333}{+} \color{#D61F06}{6} \color{#333333}{-} \color{#3D99F6}{1 } \color{#333333}{= 0}$ When the second half is written out, each digit occurs in the different colour, since $n$ has an odd number of digits. This means that 1 concatenation (2 copies of $n$ ) is always enough for each digit to cancel each other out in the alternating sum, making the alternating sum 0, which is a multiple of 11.

Case 2: $n$ has an even number of digits

For example let $n = 6859$ . If we concatenate it with itself once, we get $\huge \color{#D61F06}{6} \color{#3D99F6}{8} \color{#D61F06}{5} \color{#3D99F6}{9} \color{#D61F06}{6} \color{#3D99F6}{8} \color{#D61F06}{5} \color{#3D99F6}{9}$ $\text{Alternating Sum} = \color{#D61F06}{6} \color{#333333}{-} \color{#3D99F6}{8} \color{#333333}{+} \color{#D61F06}{5} \color{#333333}{-} \color{#3D99F6}{9} \color{#333333}{+} \color{#D61F06}{6} \color{#333333}{-} \color{#3D99F6}{8} \color{#333333}{+} \color{#D61F06}{5} \color{#333333}{-} \color{#3D99F6}{9}\color{#333333}{= -12 \equiv 10 \mod{11}}$ This time, for each extra concatenation, each digit occurs in the same colour each time. This means that the digits won't cancel themselves out now. With 0 concatenations, there will be an alternating sum, $a$ of $0 \leq a \leq 10 \mod 11$ $\huge \color{#D61F06}{6} \color{#3D99F6}{8} \color{#D61F06}{5} \color{#3D99F6}{9}$ $\text{Alternating Sum} = \color{#D61F06}{6} \color{#333333}{-} \color{#3D99F6}{8} \color{#333333}{+} \color{#D61F06}{5} \color{#333333}{-} \color{#3D99F6}{9} \color{#333333}{= -6 \equiv 5 \mod{11}}$ For each extra concatenation, the alternating sum is increased by this amount (5 in this case). Since $a$ and $11$ will be coprime, 10 concatenations (11 copies of $n$ ) will be needed to ensure that the alternating sum is divisible by 11. In our example, the alternating sum would equal $11 \times 5 = 55 \equiv 0 \mod 11$ .

As one case requires 2 copies of $n$ and the other case requires 11 copies of $n$ , to ensure that the result will be a multiple no matter how many digits $n$ has we need $\text{Lowest Common Multiple}(2,11) = \boxed{22}$ copies of $n$ .

Note that $10 \equiv -1 \pmod{11}$ , and so $10^m \equiv \pm1 \pmod{11}$ , depending on whether $m$ is even or odd. Thus $1 + 10^m + 10^{2m} + \cdots + 10^{10m} \equiv 0 \pmod{11}$ when $m$ is even, while $1 + 10^m \equiv 0 \pmod{11}$ when $m$ is odd. Thus, if $n$ has an even number of digits, then the concatenation $\overline{nnnnnnnnnnn}$ (of $11$ copies of $n$ ) is a multiple of $n$ (and, in general, the concatenation of a smaller number of copies of $n$ will not be). Also, if $n$ has an odd number of digits, then the concatenation $\overline{nn}$ of two copies of $n$ is a multiple of $11$ . Since $1 + x + x^2 + \cdots + x^{21} \; = \; (1 + x + \cdots + x^{10})(1 + x^{11}) \; = \; (1 + x)(1 + x^2 + x+4 + \cdots + x^{20})$ we see that the concatenation $\overline{nnnnnnnnnnnnnnnnnnnnnn}$ of $\boxed{22}$ copies of $n$ will always be a multiple of $11$ , no matter how many digits $n$ has (and no smaller number of repeats will always give a multiple of $11$ , although a smaller number of repeats can be found for any particular $n$ , if we know know many digits $n$ has).

Edit : The last few lines relate to a previous wording of the question.

This makes the answer $\boxed{21}$ .

Terminologically, the number $23232323$ is more normally described as the $4$ -fold concatenation of $23$ , rather than as the result of concatenating $23$ with itself $3$ times.

If n has odd digits,concatenate it 2-1=1 time.
If n has even digits,concatenate it 11-1=10 times.
So for any digits,concatenate it (2,11)-1=21 times.

The wording could be a little better. I agree with odd length requiring 2 copies and even requiring 11. Thus 11 copies will be sufficient according to my reading of the problem: You add copies of $n$ until you get a multiple of 11 and $\max{(2,11)}=11$

I suggest the wording should say final result is a multiple of 11

(I couldn't believe 11 wasn't the expected answer because I knew no number would require more than 11 copies. A couple of hours later I basically guessed 22.)

Edit: maybe it's just me. Other solvers seem to have understood the wording.

Simple,

If a number is divisible by 11, it means sum-diff(sum of odd places - sum of even places) should be a divisible of 11. For eg. 913, 9-1+3=11, divisible by 11.

Start:- For an odd digit number:

Let's take abc for now as its digits. The sum-diff will be a - b + c. Concatenation will make it abcabc and the sum will be (a-b+c) - (a-b+c) =0, divisible by 11. Similarly, for any odd digit number we need only two concatenations to get a sum 0, divisible by 11. Try- 9876598765

For an even digit number:

Let's take ab, the sum-diff will be a-b. For n concatenations it will be n(a-b). If a=b, we need no concatenations; otherwise, we need 11 concatenations for any value of a-b (that will be an integer between -8 and 9 except zero, and to get a multiple of 11 you need 11 summations of such digits, reason: 11 is a prime number). Similar is the case with any number of even digits.

Concluded: For odd number we need only 2 concatenations or any number that is divisible by 2. For even number we need 11 concatenations or any number that is divisible by 11. We would like to take LCM of 2 and 11.

It should be: 22

:)

Since the positive integer is of unknown digit length, we can split into 2 scenarios: either odd digit length or even digit length. This is very important because of the way multiples of 11 work.

To test whether a number is a multiple of 11, the alternating sum must be a multiple of 11.

If you look at a number which is odd in digit length, for example 215: 2 - 1 + 5; You will notice that if you concatenate 2 of these, the digits will cancel out and make 0: 2 - 1 + 5 - 2 + 1 - 5 = (2-2) + (-1+1) + (5-5) = 0 This pattern is noticed with ANY number with an odd digit length concatenated an even number of times. Therefore to accommodate the scenario of n having an odd digit length, the minimum number of copies must be even.

Now for the second option: if n has an even length. Because of the digit placement, the alternating sum does not cancel out nicely unlike the first scenario. If we use 21 as an example, we can look at it like this: 2 - 1 = 1; so if we concatenate it 5 times, for example, this sum adds up: 2 - 1 + 2 - 1 + 2 - 1 + 2 - 1 + 2 - 1 = 1 + 1 + 1 + 1 + 1 = 5; So to get a multiple of 11, simply concatenate the number 11 times because the digits will always add up. It doesn't matter if the alternating sum of n isn't 1, it will be required to be concatenated 11 times to produce an alternating sum that is a multiple of 11. Therefore for this scenario, the number of times the number is concatenated to create a multiple of 11 must be a multiple of 11.

Putting the two scenarios together (must be a multiple of 2 and must be a multiple of 11), this minimum number that does this is 22.

The lack of information of the length forces one to choose a number of copies that will work under all circumstances. Hence the problem is equivalent to:

What is the smallest number $N\in\mathbb{N}^{+}$ , such that for all $n\in\mathbb{N}^{+}$ the concatenation $n^{\otimes N}:=\underbrace{n_{10}\Vert n_{10}\Vert \ldots\Vert n_{10}}_{N}$ satisfies $n^{\otimes N}\equiv 0$ mod $11$ . Hereby $k_{10}$ denotes the digit expansion of $k$ to base $b=10$ . In other words, setting

$C(n):=\{N\in\mathbb{N}^{+}\mid n^{\otimes N}\equiv 0\text{ mod }b+1\}$

for all $n\in\mathbb{N}^{+}$ , what is $\min\bigcap_{n\in\mathbb{N}^{+}}C(n)$ ?

Under this formulation, one has the following approach.

First note that $b+1\in\mathbb{P}$ and that $b\equiv -1$ mod $b+1(=11)$ . Let $n\in\mathbb{N}$ be of length $L\in\mathbb{N}^{+}$ . Now, modulo $b+1$ one computes for arbitrary $N\in\mathbb{N}^{+}$

$\begin{array}{c}[mc]{rcl} n^{\otimes N} &= &\sum_{k<N}b^{kL}\cdot n\\ &\equiv &\sum_{k<N}((-1)^{L})^{k}\cdot n\\ &\equiv &\left\{ \begin{array}{c}[mc]{lcl} \sum_{k<N}1\cdot n &: &L\text{ is even}\\ \sum_{k<N}(-1)^{k}\cdot n &: &L\text{ is odd}\\ \end{array}\right.\\ &\equiv &\left\{ \begin{array}{c}[mc]{lcl} N\cdot n &: &L\text{ is even}\\ 0 &: &L\text{ is odd},~N\text{ is even}\\ n &: &L\text{ is odd},~N\text{ is odd}\\ \end{array}\right. \end{array}$

• Case 1. $n\equiv 0$ mod $b+1$ . Then $1$ copy suffices. Hence $C(n)=\mathbb{N}^{+}$ .
• Case 2. $n\not\equiv 0$ mod $b+1$ and $n$ is of odd length. Then by the above computation $n^{\otimes N}\equiv 0$ mod $b+1$ if an even number of copies is used, or else $n^{\otimes N}\equiv n\not\equiv 0$ mod $b+1$ if one uses an odd number of copies. Hence $C(n)=2\cdot\mathbb{N}^{+}$ .
• Case 2. $n\not\equiv 0$ mod $11$ and $n$ is of odd length. Then by the above computation, $n^{\otimes N}\equiv Nn$ mod $b+1$ . Since $b+1\in\mathbb{P}$ and $b+1$ does not divide $n$ , we have $n^{\otimes N}\equiv 0$ if and only if $N\equiv 0$ mod $b+1$ . Hence $C(n)=(b+1)\cdot\mathbb{N}^{+}$ .

Now here comes the crux. If we knew what $n$ was, then the answer would be $\min C(n)$ , which is either $1$ , $2$ or $b+1=11$ depending upon $n$ . The answer to the problem as stated (given the ‘unknowledge’) is

$\begin{array}{c}[mc]{rcl} \min\bigcap_{n\in\mathbb{N}^{+}}C(n) &= &\min(\mathbb{N}^{+}\cap 2\cdot\mathbb{N}^{+}\cap (b+1)\cdot\mathbb{N}^{+})\\ &= &lcm(1,2,(b+1))=2(b+1)=22\\ \end{array}$

since $2,b+1\in\mathbb{P}$ and $b+1\geq 2+1=3>2$ . Hence, $\fbox{22}$ is the minimum number of copies that will always work under all circumstances.

Let $d=10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+10^2.a_2+10.a_1+a_0$ be a $n$ digit number. Also we know that $10^m \equiv \begin{cases} +1 & m \equiv 2k\\-1 & m \equiv (2k+1) \end{cases}$ Now for $d$ to be $0 \mod 11$ $n$ must be odd because if $n$ is even then the magnitude of difference of the sum of digits in the odd places and the even places increases without bounds no matter whatever the value of $n$ may be. Note that in this case the digits in the odd places and that in the even places hold their position after each concatenation. $$ SO that being said now for odd $n$ Let $d$ be $abcde$ for ex; now after 1st concatenation we have $abcdeabcde$ Note that if the place of a digit $i$ in $d$ is $10^x$ then after $d$ being concatenated with itself $i$ occurs additionally at $10^{x+n}$ . Since $n$ is odd so the index of $10$ corresponding to the two value of $i$ have odd parity!!! So after any odd number of concatenations we have an even number of $i$ with us half of which yields $+i \mod 11$ and the other half yields $-i \mod 11$ for all $i$ ; That solves the problem for this case. $$ In the second case when n is even then we need exactly 11 concatenations for the magnitude of the difference of sum of digits in the odd and even places to be a multiple of $11$ . So combining the previous two cases we get our answer to be $LCM(case 1,case 2) = 22$ .