Linked Questions

37
votes
5answers
6k views

Show that there are infinitely many powers of two starting with the digit 7 [duplicate]

This is a contest math problem which I was not able to solve. A hint toward the solution would be helpful as well. Problem: Show that there are infinitely many powers of 2 starting with the digit 7. ...
4
votes
1answer
712 views

Is $2^k = 2013…$ for some $k$? [duplicate]

I'm wondering if some power of $2$ can be written in base $10$ as $2013$ followed by other digits. Formally, does there exist $k,q,r \in \mathbb N$ such that $$2^k=2013 \cdot 10^q+r \,\,\,; \,\,\,r&...
6
votes
1answer
617 views

Is it known or new? [duplicate]

Possible Duplicate: Starting digits of 2^n While I was randomly working with number patterns, I came along with some interesting pattern which seems to turn to a conjecture in fact. My ...
4
votes
2answers
170 views

Does $\exists n \in \mathbb{Z}$ such that $2^n$ can start with $9786543120$? [duplicate]

I have tried $2^n=\displaystyle\sum_{k=0}^{n}\binom{n}{k}$ but could not reach further. Thank you.
3
votes
2answers
279 views

Decimal expansion of $2^n$ [duplicate]

i am trying to read / understand the book: Introduction to the modern theory of dynamic systems, hassleblatt boris and katok anatole. On page 28 there is an excercise which I am trying to solve but ...
5
votes
1answer
105 views

For any $n$ there's a power of $2$ which contains $n$ [duplicate]

So I saw this problem in an Olympiad book, "Prove that for any natural number $n$, there exists a power of $2$ which contains $n$ in it. " For example, $n=19$ is in $2^{13}=8192$, $n=24$ is in $2^{...
2
votes
1answer
72 views

Left most Digits of $5^n$ [duplicate]

Prove that for every integer $m$ there is an integer $n$ such that the digits of $5^n$ start with $m$ (left most digits). For example for $m=156$, $n=6$ is the solution, because $5^6 = \color{red}{156}...
4
votes
0answers
100 views

problem about number and power of 2 [duplicate]

Possible Duplicate: Starting digits of 2^n can anyone give me a hint ? Prove that any finite sequence of digits is a starting sequence of digits for some power of $2$. This is my attempt : ...
5
votes
0answers
63 views

Proving the existence of an integer $n$ such that $2^n=123456789…$ [duplicate]

How to prove that it exists an integer $n$ such that the decimal expansion of $2^n$ starts with $123456789$?
1
vote
1answer
45 views

Consider the sequence $1,2,4,8,…,a^n = 2^n,…$ of all the powers of $2$ [duplicate]

Consider the sequence $1,2,4,8,...,a^n = 2^n,...$ of all the powers of $2$. Prove that, given any digit $i ∈ {1,...,9}$, there exist infinitely many values of $n$ for which $a^n$ starts with that ...
1
vote
0answers
42 views

Given $n$, exist $k$ such that $2^k$ contains $n$ as string. [duplicate]

I have this doubt: Given $n\in \mathbb{N}$, does exist $k\in \mathbb{N}$ such that $2^k$ contains $n$ as a string in it? For example, $53$ is in $2^{16}=65\color{red}{53}6$. I just thought the ...
10
votes
1answer
1k views

Do the Fibonacci numbers contain any run of digits?

Related to this question and inspired by this challenge on PPCG. The challenge is as follows: for a given natural number $x \in \mathbb{N}$, find the first Fibonacci number $F_n$ that contains $x$. ...
5
votes
3answers
834 views

Fractional part of $b \log a$

From the problem... Find the minimal positive integer $b$ such that the first digits of $2^b$ are 2011 ...I have been able to reduce the problem to the following instead: Find minimal $b$ such ...
4
votes
2answers
691 views

Group theory with analysis

I've studied group theory upto isomorphism. Topics include : Lagrange's theorem, Normal subgroups, Quotient groups, Isomorphism theorems. I too have done metric spaces and real analysis properly. ...
0
votes
3answers
212 views

Is it possible that $3^n$ starts with $2019$ for some positive integer? [duplicate]

Does there exist a positive integer $n$ such that the decimal representation of $3^n$ starts with $2019$? My attempt: I have tried some starting powers and I guess no such positive integers exists. ...

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