1. ## Prove:

For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1

2. Originally Posted by newtoinequality
For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
Hi

It can be done by induction
The property is true for n=2 (16 = 2^4)

Suppose that it is true for n. This means that there exists p integer such that $\displaystyle 10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}$

You have to demonstrate that there exists q integer such that $\displaystyle 10^{n} \leq 2^q < 2 \cdot 10^{n}$

3. Originally Posted by running-gag
Hi

It can be done by induction
The property is true for n=2 (16 = 2^4)

Suppose that it is true for n. This means that there exists p integer such that $\displaystyle 10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}$

You have to demonstrate that there exists q integer such that $\displaystyle 10^{n} \leq 2^q < 2 \cdot 10^{n}$
You mean $\displaystyle 10^n\leq 2^{n+1}< 2\cdot 10^{n}$, don't you?

4. Originally Posted by newtoinequality
For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
First, show existence. If the statement were false, then there would exist $\displaystyle m,\,n$ such that $\displaystyle 2^m<10^{n-1}$ and $\displaystyle 2^{m+1}\ge2\cdot10^{n-1}.$ But the former gives $\displaystyle 2^{m+1}<2\cdot10^{n-1}$ which is a contradiction of the latter.

Hence for each $\displaystyle n$ there must be $\displaystyle m$ such that $\displaystyle 10^{n-1}\le2^m<2\cdot10^{n-1},$ i.e.

$\displaystyle 2^m\ <\ 2\cdot10^{n-1}\quad\ldots\ \fbox1$

$\displaystyle 2^m\ \ge\ 10^{n-1}\quad\ldots\ \fbox2$

To prove uniqueness, observe that $\displaystyle 2^{m-1}<10^{n-1}$ (from $\displaystyle \fbox1)$ and $\displaystyle 2^{m+1}\ge2\cdot10^{n-1}$ (from $\displaystyle \fbox2).$ The result follows since if $\displaystyle m'\ne m$ then either $\displaystyle 2^{m'}\le2^{m-1}$ or $\displaystyle 2^{m'}\ge2^{m+1}$ and so $\displaystyle 2^{m'}$ cannot be between $\displaystyle 10^{n-1}$ and $\displaystyle 2\cdot10^{n-1}.$

5. Originally Posted by HallsofIvy
You mean $\displaystyle 10^n\leq 2^{n+1}< 2\cdot 10^{n}$, don't you?
No

For instance $\displaystyle 10^3 \leq 2^{10} < 2 \cdot 10^3$