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

Is $2^k = 2013…$ for some $k$?Is there n such that $3^n$ starts with 2019Starting digits of $2^n$.Combinatorics - Show that $2^n = 1987 dots $Numbers with integer multiples using only digits $2$ and $6$ (Austria Mathematical Olympiad 2006)Prove or disprove that there exists a unique positive integer sequence ${a_{n}}$ satisfying a conditionNumbers $n$ such that there is some digit occuring in each power of $n$Decimal expansion of $kn$ contains only $0,7$ for some $k in mathbb{N}$Unique representations of positive integers?Prove that there exist $n$ consecutive positive integers with the $i$th integer divisible by the $i$th primeIf $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.Decimal representation begins with $N$Show that there do not exist any integer $m,n$ such that $frac{m}{n}+frac{n+1}{m}=4$Extremal problem with positive integer numbers

2

2

$begingroup$

This question already has an answer here:

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

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. Apart from this I have tried binomial expansion $3^n = (1+2)^n = 1
+ {nchoose 1}2 + {n choose 2}2^2 + cdots+ {n choose n }2^n$ but I don't no how to proceed here. Any hint will be helpful.

abstract-algebra number-theory

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edited Mar 11 at 15:13

Rócherz

2,9762821

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

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marked as duplicate by Dietrich Burde abstract-algebra
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Mar 11 at 12:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is an old question, see here, or here, or here etc.
  $endgroup$
  – Dietrich Burde
  Mar 11 at 12:13
  
  
  

  
  
  
  

add a comment | 

2

2

$begingroup$

This question already has an answer here:

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

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. Apart from this I have tried binomial expansion $3^n = (1+2)^n = 1
+ {nchoose 1}2 + {n choose 2}2^2 + cdots+ {n choose n }2^n$ but I don't no how to proceed here. Any hint will be helpful.

abstract-algebra number-theory

share|cite|improve this question

edited Mar 11 at 15:13

Rócherz

2,9762821

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

$endgroup$

marked as duplicate by Dietrich Burde abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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Mar 11 at 12:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is an old question, see here, or here, or here etc.
  $endgroup$
  – Dietrich Burde
  Mar 11 at 12:13
  
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

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

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. Apart from this I have tried binomial expansion $3^n = (1+2)^n = 1
+ {nchoose 1}2 + {n choose 2}2^2 + cdots+ {n choose n }2^n$ but I don't no how to proceed here. Any hint will be helpful.

abstract-algebra number-theory

share|cite|improve this question

edited Mar 11 at 15:13

Rócherz

2,9762821

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

$endgroup$

share|cite|improve this question

edited Mar 11 at 15:13

Rócherz

2,9762821

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

share|cite|improve this question

edited Mar 11 at 15:13

Rócherz

2,9762821

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

edited Mar 11 at 15:13

Rócherz

2,9762821

edited Mar 11 at 15:13

Rócherz

2,9762821

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

asked Mar 11 at 11:53

CHOUDHARY bhim senCHOUDHARY bhim sen

1589

3 Answers
3

active

oldest

votes

5

$begingroup$

You want positive integers $m$ and $n$ such that
$$
2019 cdot 10^mleq3^n<2020cdot 10^m.
$$
Taking the $log$ each each side gives $$
log(2019)+mleq n log(3)<log(2020)+m \ implies frac{log(2019)}{log(3)}+frac{m}{log(3)}leq n<frac{log(2020)}{log(3)}+frac{m}{log(3)}.$$
A positive integer lies in the interval $$
left(frac{log(2019)}{log(3)}+frac{m}{log(3)},frac{log(2020)}{log(3)}+frac{m}{log(3)}right)$$
if $$left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}>1-frac{log(2020)-log(2019)}{log(3)}$$ where ${x}=x-[x]$. But since $frac 1{log(3)}$ is irrational the set $left{left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}:minmathbb Nright}$ is dense in $[0,1)$. Thus such an $n$ indeed exists.

share|cite|improve this answer

edited Mar 11 at 13:27

Rócherz

2,9762821

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

$endgroup$

add a comment | 

3

$begingroup$

Since $$log_{10}(3)cdot n$$ is equidistributed modulo $1$ since $log_{10}(3)$ is irrational, in fact $3^n$ can start with every given finite digit string with first digit non-zero.

Brute force reveals that $3^{686}$ is the smallest power of $3$ starting with $2019$

share|cite|improve this answer

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $n|log3-frac{m}{n}|<log(2020/2019)approx0.0005$, you can narrow the search by finding continued fraction approximations that give this amount of accuracy. In this case $log3approxfrac{713}{649}pm0.0006$ gives 649 as a good starting point for the search.
  $endgroup$
  – Chrystomath
  Mar 11 at 12:39
  

  
  
  
  

add a comment | 

1

$begingroup$

Yes, in fact there are infinitely many $ninmathbb{N}$ such that $3^n$ starts with $2019$.

$3^n$ starts with $2019$ if and only if $$2019 cdot 10^m le 3^n < 2020 cdot 10^m$$
for some $m in mathbb{N}_0$. This is equivalent to
$$m + log_{10} 2019 le nlog_{10}3 < m + log_{10}2020$$

Now, since $log_{10}3$ is irrational, the set $${nlog_{10}3 - lfloor nlog_{10}3rfloor : n inmathbb{N}}$$ is dense in $[0,1]$ so there exists infinitely many $n in mathbb{N}$ such that
$$nlog_{10}3 - lfloor nlog_{10}3rfloor in [log_{10}2019 - 3, log_{10}2020 - 3) subseteq [0,1]$$

or equivalently
$$(lfloor nlog_{10}3rfloor-3) + log_{10}2019 le nlog_{10}3 < (lfloor nlog_{10}3rfloor-3) + log_{10}2020 $$

share|cite|improve this answer

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548

$endgroup$

add a comment | 

5

$begingroup$

You want positive integers $m$ and $n$ such that
$$
2019 cdot 10^mleq3^n<2020cdot 10^m.
$$
Taking the $log$ each each side gives $$
log(2019)+mleq n log(3)<log(2020)+m \ implies frac{log(2019)}{log(3)}+frac{m}{log(3)}leq n<frac{log(2020)}{log(3)}+frac{m}{log(3)}.$$
A positive integer lies in the interval $$
left(frac{log(2019)}{log(3)}+frac{m}{log(3)},frac{log(2020)}{log(3)}+frac{m}{log(3)}right)$$
if $$left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}>1-frac{log(2020)-log(2019)}{log(3)}$$ where ${x}=x-[x]$. But since $frac 1{log(3)}$ is irrational the set $left{left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}:minmathbb Nright}$ is dense in $[0,1)$. Thus such an $n$ indeed exists.

share|cite|improve this answer

edited Mar 11 at 13:27

Rócherz

2,9762821

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

$endgroup$

add a comment | 

5

$begingroup$

You want positive integers $m$ and $n$ such that
$$
2019 cdot 10^mleq3^n<2020cdot 10^m.
$$
Taking the $log$ each each side gives $$
log(2019)+mleq n log(3)<log(2020)+m \ implies frac{log(2019)}{log(3)}+frac{m}{log(3)}leq n<frac{log(2020)}{log(3)}+frac{m}{log(3)}.$$
A positive integer lies in the interval $$
left(frac{log(2019)}{log(3)}+frac{m}{log(3)},frac{log(2020)}{log(3)}+frac{m}{log(3)}right)$$
if $$left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}>1-frac{log(2020)-log(2019)}{log(3)}$$ where ${x}=x-[x]$. But since $frac 1{log(3)}$ is irrational the set $left{left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}:minmathbb Nright}$ is dense in $[0,1)$. Thus such an $n$ indeed exists.

share|cite|improve this answer

edited Mar 11 at 13:27

Rócherz

2,9762821

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

$endgroup$

add a comment | 

$begingroup$

You want positive integers $m$ and $n$ such that
$$
2019 cdot 10^mleq3^n<2020cdot 10^m.
$$
Taking the $log$ each each side gives $$
log(2019)+mleq n log(3)<log(2020)+m \ implies frac{log(2019)}{log(3)}+frac{m}{log(3)}leq n<frac{log(2020)}{log(3)}+frac{m}{log(3)}.$$
A positive integer lies in the interval $$
left(frac{log(2019)}{log(3)}+frac{m}{log(3)},frac{log(2020)}{log(3)}+frac{m}{log(3)}right)$$
if $$left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}>1-frac{log(2020)-log(2019)}{log(3)}$$ where ${x}=x-[x]$. But since $frac 1{log(3)}$ is irrational the set $left{left{frac{log(2019)}{log(3)}+frac{m}{log(3)}right}:minmathbb Nright}$ is dense in $[0,1)$. Thus such an $n$ indeed exists.

share|cite|improve this answer

edited Mar 11 at 13:27

Rócherz

2,9762821

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

$endgroup$

share|cite|improve this answer

edited Mar 11 at 13:27

Rócherz

2,9762821

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

edited Mar 11 at 13:27

Rócherz

2,9762821

edited Mar 11 at 13:27

Rócherz

2,9762821

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

answered Mar 11 at 12:05

Soumik GhoshSoumik Ghosh

760111

3

$begingroup$

Since $$log_{10}(3)cdot n$$ is equidistributed modulo $1$ since $log_{10}(3)$ is irrational, in fact $3^n$ can start with every given finite digit string with first digit non-zero.

Brute force reveals that $3^{686}$ is the smallest power of $3$ starting with $2019$

share|cite|improve this answer

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $n|log3-frac{m}{n}|<log(2020/2019)approx0.0005$, you can narrow the search by finding continued fraction approximations that give this amount of accuracy. In this case $log3approxfrac{713}{649}pm0.0006$ gives 649 as a good starting point for the search.
  $endgroup$
  – Chrystomath
  Mar 11 at 12:39
  

  
  
  
  

add a comment | 

3

$begingroup$

Since $$log_{10}(3)cdot n$$ is equidistributed modulo $1$ since $log_{10}(3)$ is irrational, in fact $3^n$ can start with every given finite digit string with first digit non-zero.

Brute force reveals that $3^{686}$ is the smallest power of $3$ starting with $2019$

share|cite|improve this answer

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $n|log3-frac{m}{n}|<log(2020/2019)approx0.0005$, you can narrow the search by finding continued fraction approximations that give this amount of accuracy. In this case $log3approxfrac{713}{649}pm0.0006$ gives 649 as a good starting point for the search.
  $endgroup$
  – Chrystomath
  Mar 11 at 12:39
  

  
  
  
  

add a comment | 

$begingroup$

Since $$log_{10}(3)cdot n$$ is equidistributed modulo $1$ since $log_{10}(3)$ is irrational, in fact $3^n$ can start with every given finite digit string with first digit non-zero.

Brute force reveals that $3^{686}$ is the smallest power of $3$ starting with $2019$

share|cite|improve this answer

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

$endgroup$

share|cite|improve this answer

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

answered Mar 11 at 11:56

PeterPeter

48.8k1139136

1

$begingroup$

Yes, in fact there are infinitely many $ninmathbb{N}$ such that $3^n$ starts with $2019$.

$3^n$ starts with $2019$ if and only if $$2019 cdot 10^m le 3^n < 2020 cdot 10^m$$
for some $m in mathbb{N}_0$. This is equivalent to
$$m + log_{10} 2019 le nlog_{10}3 < m + log_{10}2020$$

Now, since $log_{10}3$ is irrational, the set $${nlog_{10}3 - lfloor nlog_{10}3rfloor : n inmathbb{N}}$$ is dense in $[0,1]$ so there exists infinitely many $n in mathbb{N}$ such that
$$nlog_{10}3 - lfloor nlog_{10}3rfloor in [log_{10}2019 - 3, log_{10}2020 - 3) subseteq [0,1]$$

or equivalently
$$(lfloor nlog_{10}3rfloor-3) + log_{10}2019 le nlog_{10}3 < (lfloor nlog_{10}3rfloor-3) + log_{10}2020 $$

share|cite|improve this answer

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548

$endgroup$

add a comment | 

1

$begingroup$

Yes, in fact there are infinitely many $ninmathbb{N}$ such that $3^n$ starts with $2019$.

$3^n$ starts with $2019$ if and only if $$2019 cdot 10^m le 3^n < 2020 cdot 10^m$$
for some $m in mathbb{N}_0$. This is equivalent to
$$m + log_{10} 2019 le nlog_{10}3 < m + log_{10}2020$$

Now, since $log_{10}3$ is irrational, the set $${nlog_{10}3 - lfloor nlog_{10}3rfloor : n inmathbb{N}}$$ is dense in $[0,1]$ so there exists infinitely many $n in mathbb{N}$ such that
$$nlog_{10}3 - lfloor nlog_{10}3rfloor in [log_{10}2019 - 3, log_{10}2020 - 3) subseteq [0,1]$$

or equivalently
$$(lfloor nlog_{10}3rfloor-3) + log_{10}2019 le nlog_{10}3 < (lfloor nlog_{10}3rfloor-3) + log_{10}2020 $$

share|cite|improve this answer

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548

$endgroup$

add a comment | 

$begingroup$

Yes, in fact there are infinitely many $ninmathbb{N}$ such that $3^n$ starts with $2019$.

$3^n$ starts with $2019$ if and only if $$2019 cdot 10^m le 3^n < 2020 cdot 10^m$$
for some $m in mathbb{N}_0$. This is equivalent to
$$m + log_{10} 2019 le nlog_{10}3 < m + log_{10}2020$$

Now, since $log_{10}3$ is irrational, the set $${nlog_{10}3 - lfloor nlog_{10}3rfloor : n inmathbb{N}}$$ is dense in $[0,1]$ so there exists infinitely many $n in mathbb{N}$ such that
$$nlog_{10}3 - lfloor nlog_{10}3rfloor in [log_{10}2019 - 3, log_{10}2020 - 3) subseteq [0,1]$$

or equivalently
$$(lfloor nlog_{10}3rfloor-3) + log_{10}2019 le nlog_{10}3 < (lfloor nlog_{10}3rfloor-3) + log_{10}2020 $$

share|cite|improve this answer

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548

$endgroup$

share|cite|improve this answer

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548

answered Mar 11 at 12:11

mechanodroidmechanodroid

28.7k62548