Binary multiples

Define

$c_1 = 1$ and $c_n = 10^{n-1} + c_{n-1}$

I.e., $c_n$ is the number written with $n$ ones in a row.

Let us fix some natural $k$ and consider the sets

$S^k_i = \{ c_{1 + ik}, c_{2 + ik}, \cdots, c_{k+ik} \}, i \geq 0$

For each $S^k_i$ we will build a binary multiple of $k$ , thus proving there is an infinite amount of such binary multiples, given that there is an infinite amount of the $S^k_i$ for each $k$ .

Consider one of the $S^k_i$ . If $S^k_i$ has a multiple of $k$ , that multiple is automatically a binary multiple because of how the $c_t$ were built. If none of its elements is a multiple, then by the pigeonhole principle two of its elements, say $c_{a+ik} > c_{b+ik}$ have the same remainder when divided by $k$ which in turn implies $c_{a+ik} - c_{b + ik}$ is a binary multiple of $k$ .

It is straightforward to verify that each $S^k_i$ produces a different binary multiple of $k$ when $k$ is fixed, proving there is an infinite amount of them.

Let $k$ be a natural number. Then the decimal representation of $\frac{1}{k}$ either terminates or repeats.

Case 1: The decimal representation of $\frac{1}{k}$ terminates.

Then the decimal can be written as a fraction $\frac{p}{10^q}$ for some positive integers $p$ and $q$ , and since $\frac{1}{k} = \frac{p}{10^q}$ , $kp = 10^q$ , making $10^q$ a binary multiple of $k$ .

For example, if $k = 8$ , then since $\frac{1}{8} = 0.125 = \frac{125}{1000}$ , $8 \cdot 125 = 1000$ making $1000$ a binary multiple of $8$ .

Case 2: The decimal representation of $\frac{1}{k}$ repeats.

Then the decimal can be written as terminating decimal (which can be written as a fraction $\frac{p}{10^q}$ for some positive integers $p$ and $q$ ) plus an infinite sum (which using a starting term of $t_1 = \frac{s}{10^u}$ and a ratio of $r = \frac{1}{10^v}$ for some positive integers $s$ , $u$ , and $v$ where $u > v$ and the infinite sum equation $S = \frac{t_1}{1 - r}$ can be written as a fraction $\frac{s}{10^{u - v}(10^v - 1)}$ ), which means $\frac{1}{k} = \frac{p}{10^q} + \frac{s}{10^{u - v}(10^v - 1)} = \frac{w}{10^x(10^v - 1)}$ for $w = 10^{u - v}(10^v - 1)p + 10^qs$ and $x = u - v + q$ , and so $kw = 10^x(10^v - 1)$ .

Now, $10^x(10^v - 1)$ is a string of $v$ $9$ s followed by a string of $x$ $0$ s in decimal notation, which divides evenly into $y$ , a string of $9v$ $1$ s followed by a string of $x$ $0$ s, because if $L_n$ is a string of $n$ $1$ s then $\frac{y}{10^x} = L_{9v}$ $=$ $L_v(10^{8v} - 1) + L_v(10^{7v} - 1) + L_v(10^{6v} - 1)$ $+$ $L_v(10^{5v} - 1) + L_v(10^{4v} - 1) + L_v(10^{3v} - 1)$ $+$ $L_v(10^{2v} - 1) + L_v(10^{v} - 1) + 9L_v$ , and since each term is divisible by $10^v - 1$ , $\frac{y}{10^x}$ is a multiple of $10^v - 1$ , which means $y$ is a binary multiple of $10^x(10^v - 1)$ .

Therefore, since $y$ is a binary multiple of $10^x(10^v - 1)$ and $10^x(10^v - 1)$ is a multiple of $k$ , $v$ is a binary multiple of $k$ .

For example, if $k = 12$ , then since $\frac{1}{12} = 0.08333.. = \frac{8}{100} + \frac{3}{900} = \frac{75}{900}$ , $12 \cdot 75 = 900$ , and since $900$ divides evenly into $11111111100$ , $11111111100$ is a binary multiple of $900$ and therefore also a binary multiple of $12$ .

Therefore, from both cases we can conclude that every natural number k has at least one binary multiple. But we can also show that if at least one binary multiple exists, infinitely many binary multiples must exist, because if $k$ is a natural number and $n$ is a binary multiple of $k$ , then by definition $\frac{n}{k}$ is an integer, which means $\frac{10^an}{k}$ is also an integer for any integer $a$ , and $k \cdot \frac{10^an}{k} = 10^an$ , making $10^an$ a binary multiple of $k$ for any integer $a$ .

Therefore, every natural number has an infinite amount of binary multiples .

First we can see that any binary-multiple can be expressed as the sum of distinct powers of 10, for example $10101=10^4+10^2+10^0$ .

Let $n$ be a natural number.

We can categorize the first $n^2$ powers of 10 by their modulo $n$ . Then, by the pidgeonhole principle. There's at least one category with $\lceil \frac{n^2}{n} \rceil = n$ or more elements.

So if we sum $n$ powers of 10 from the $x$ category, we'll obtain a binary-multiple $k$ that satisfy: $k \equiv nx \equiv 0x \equiv 0 \pmod{n}$ In other words, $k$ is a multiple of $n$ .

Also $10k, 100k,...$ , are binary-multiples of $n$ .

So every natural number $n$ has infinite number of binary-multiples.

Take any positive k you like. By the pidgeonhole prinicple, there exists some pair of repunits which have the same value mod k. The difference between any two repunits is another one followed by zeros, which can qualify as a binary multiple. Since the two repunits selected are equal mod k, their difference is 0 mod k, and the number is a binary multiple.

To show every k has infinitely many, you could show how the pidgeonhole principle can be extended to give infinite pairs of repunits. However, it is simpler to just note that 10 times a binary multiple of k is also a binary multiple of k. This can be repeated indefinitely, so every k has infinite binary multiples.