Powers of $2$ 's After Another

Relevant wiki: Induction - Introduction

By induction. Let $A(n)$ be the number generated by concatenating the powers $2^0, \dots, 2^n$ .

Because obviously $2^0|A(0)$ , the basis step for induction is satisfied.

To prove that $2^n|A(n)$ holds true for all $n \geq 0$ , suppose $2^{n-1} | A(n-1)$ . In the induction step we must prove that then also $2^n|A(n)$ .

Let $d \geq 1$ be the number of digits in $2^n$ . Then $A(n) = A(n-1)\:10^d + 2^n.$ Since $2|10^d$ and $2^{n-1}|A(n-1)$ , the first term is a multiple of $2^n$ . The same it obviously true for the second term, so that $2^n|A(n)$ .

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Generalization

In general, the number obtained by concatenating the powers $b^0, b^1, \dots, b^n$ are all divisible by $b^n$ if $b | B$ , where $B$ is the base of the number system in which we work. The proof is a generalization of the proof given above, observing that $A(n) = A(n-1)\:B^d + b^n$ is divisible by $b^n$ .

We note that $N$ can be expressed as

$\begin{aligned} N(0) & = 1 \\ N(1) & = 1\times 10^1 + 2 \times 10^0 = 12 \\ N(2) & = 2^0\times 10^2 + 2^1 \times 10^1 + 2^2 \times 10^0 = 124 \\ N(3) & = 2^0\times 10^3 + 2^1 \times 10^2 + 2^2 \times 10^1 + 2^3 \times 10^0 = 1248 \\ N(4) & = 10 \left(2^0\times 10^4 + 2^1 \times 10^3 + 2^2 \times 10^2 + 2^3 \times 10^1\right) + 2^4 \times 10^0 = 124816 \\ & = 2^0\times 10^{4\color{#D61F06}+1} + 2^1 \times 10^{3\color{#D61F06}+1} + 2^2 \times 10^{2\color{#D61F06}+1} + 2^3 \times 10^{1\color{#D61F06}+1} + 2^4 \times 10^0 = 124816 \\ \implies N(n) & = \sum_{k=0}^n 2^k\times 10^{n-k+\color{#D61F06}m_k} & \small \color{#D61F06} \text{where } m_k \text{ is a non-negative integer.} \\ & = \sum_{k=0}^n 2^{n+m_k}\times 5^{n-k+m_k} \\ & = 2^n \sum_{k=0}^n 2^{m_k}\times 5^{n-k+m_k} \\ \frac N{2^n} & = \sum_{k=0}^n 2^{m_k}\times 5^{n-k+m_k} \end{aligned}$

$\implies$ Yes, it is true. $2^n \mid N$ .

Let $N(n)$ denote as the concatenation of $2^0,2^1,2^2,…,2^n$ . For example, $N(4)=124816$ . $N(n)=2^n+2^{n-1}×10^{a_{n-1}}+2^{n-2}×10^{a_{n-2}}+…+2^1×10^{a_{1}}+2^0+10^{a_{0}}$ The $a_{k}$ control the position of $2^k$ in $N(n)$ . Now, we gonna prove that for every non-negative integer $k\leq n$ , $2^n|(2^k×10^{a_{k}})$ . Obviously $a_{n-1}\geq 1$ and $a_{0}>a_{1}>a_{2}>…>a_{n-2}>a_{n-1}$ . This means that $a_{n-t} \geq t$ ( $t={1,2,3,…n}$ ). Let $k=n-t$ , we can get $a_{k} \geq n-k$ . Let $f(a)$ as the largest power of $2$ in $a$ .(exp: $f(4)=2,f(24)=3$ )

$\begin{aligned} f(2^k×10^{a_{k}}) &\geq f(2^k×10^{n-k}) \\ &=f(2^k×2^{n-k}×5^{n-k}) \\ &=n \end{aligned}$ Hence, $N(n)$ is a sum of multiple of $2^n$ , so it is $\large true$ for any non-negative integer $n$ , $2^n|N(n)$ .

Let $f(n) = (((((2^{0}||2^{1})||2^{2})||2^{3}) \cdots )||2^{n})$ where $n \in N^+$ .

When $n = 1$ ,

$f(1) = 2^0||2^1 = 12$ and $f(1) \equiv 0\pmod{2^1}$

Now assume that when $n = k$ ,

$f(k) = 0\pmod{2^k}$

It follows that if $n = k + 1$ ,

$f(k + 1) = f(k)||2^{k + 1}$

For two integers in base 10 $(a, b)$ , $a||b = a\times10^{\lfloor log_{10}{b}\rfloor + 1} + b$

$\therefore f(k + 1) = f(k)\times10^{\lfloor log_{10}{2^{k + 1}}\rfloor + 1} + 2^{k+1}$

Now equating $\dfrac{f(k + 1)}{2^{k + 1}}$ ,

$\dfrac{f(k + 1)}{2^{k + 1}} = \dfrac{f(k)\times10^{\lfloor log_{10}{2^{k + 1}}\rfloor + 1} + 2^{k+1}}{2^{k + 1}}$

$\rightarrow \dfrac{f(k)\times10^{\lfloor (k + 1)log_{10}{2}\rfloor + 1}}{2^{k + 1}} + 1 \rightarrow \dfrac{f(k)\times10^{\lfloor (k + 1)log_{10}{2}\rfloor + 1}}{2\times2^{k}} + 1$

Since $\lfloor x \rfloor \in N^+$ for any real number $x$ , let $\varphi = \lfloor (k + 1)log_{10}{2}\rfloor$

It follows that,

$\rightarrow \dfrac{f(k)\times10^{\varphi + 1}}{2\times2^{k}} + 1$

Since we assumed that $f(k) = 0\pmod{2^k}$ , we can let $\alpha = \dfrac{f(k)}{2^k}$ .

$\therefore \dfrac{f(k+1)}{2^{k + 1}} = \dfrac{\alpha\times10^{\varphi + 1}}{2} + 1$ .

Finally, since $10^n \equiv 0\pmod2$ , $f(n) \equiv 0\pmod{2^n}$ so $\boxed{\text{Yes}}$ the statement above is true for $n \in N^+$ .

n+1 = 2(n).......... 3rd column -> n+1 divided by n is multiple of 2, means, n is a factor of n+1 ............ So "YES"

We may define the sequence by the following recursive relation: $S_0 = 2^0, S_{n+1} = 10^{\lceil \log_{10} (2^n) \rceil }S_n + 2^{n+1}$ and the problem may be rephrased as proving that $S_n \equiv 0 \mod 2^n$ . I proceed by induction, the picture in the problem already gives us a base case so for the inductive case suppose that indeed $S_n \equiv 0 \mod 2^n$ . Then $S_{n+1} \equiv 10^{\lceil \log_{10} (2^n) \rceil }S_n + 2^{n+1} \equiv 10^{\lceil \log_{10} (2^n) \rceil }S_n \equiv 0 \mod 2^{n+1}$ because $S_n$ already has a factor of $2^n$ and then the $10$ gives it at least an extra factor of $2$ to make it divisible by $2^{n+1}$ .

here is the OEIS link

Starting from n=5, Any concatenated number can be represented as or broken down to (10^y * 2^n-1 * x), which is always divisible by 2^n. Because until n=4 we know that the concatenated number is divisible by 2^n. 10^y results from subtracting 2^n from concatenated number. 2^n-1 * x is a result of the fact that previous concatenated number is divisible by 2^n-1 and this series continues....