Find numbers with n-divisors in a given range

0

$begingroup$

I'm trying to answer this question.

Are there positive integers $le200$ which have exactly 13 positive divisors? What about 14 divisors? If yes, write them. If no, explain why not.

Because I'm only familiar with prime factorization, the sieve of Eratosthenes etc. about this topic I need help to understand how to reasoning with this kind of questions.

prime-numbers divisibility prime-factorization

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asked Jan 17 at 11:44

PCNFPCNF

1338

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you count repeated divisors? Like $2^3$ ?
  $endgroup$
  – Matti P.
  Jan 17 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Yes, I suppose
  $endgroup$
  – PCNF
  Jan 17 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I guess a number with 13 positive divisors can be written as the product $$ p_1 p_2 p_3 ldots p_{12}p_{13} $$ where $p_i$ are prime numbers, not necessarily different ones. So then the smallest of such number would be obtained by choosing $p_i=2$, resulting in $2^{13}>200$.
  $endgroup$
  – Matti P.
  Jan 17 at 12:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PCNF are you familiar with $tau(n)$ function ?
  $endgroup$
  – user626177
  Jan 17 at 12:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @someone only heard but never studied or used
  $endgroup$
  – PCNF
  Jan 17 at 14:10
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm trying to answer this question.

Are there positive integers $le200$ which have exactly 13 positive divisors? What about 14 divisors? If yes, write them. If no, explain why not.

Because I'm only familiar with prime factorization, the sieve of Eratosthenes etc. about this topic I need help to understand how to reasoning with this kind of questions.

prime-numbers divisibility prime-factorization

share|cite|improve this question

asked Jan 17 at 11:44

PCNFPCNF

1338

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you count repeated divisors? Like $2^3$ ?
  $endgroup$
  – Matti P.
  Jan 17 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Yes, I suppose
  $endgroup$
  – PCNF
  Jan 17 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I guess a number with 13 positive divisors can be written as the product $$ p_1 p_2 p_3 ldots p_{12}p_{13} $$ where $p_i$ are prime numbers, not necessarily different ones. So then the smallest of such number would be obtained by choosing $p_i=2$, resulting in $2^{13}>200$.
  $endgroup$
  – Matti P.
  Jan 17 at 12:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PCNF are you familiar with $tau(n)$ function ?
  $endgroup$
  – user626177
  Jan 17 at 12:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @someone only heard but never studied or used
  $endgroup$
  – PCNF
  Jan 17 at 14:10
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

I'm trying to answer this question.

Are there positive integers $le200$ which have exactly 13 positive divisors? What about 14 divisors? If yes, write them. If no, explain why not.

Because I'm only familiar with prime factorization, the sieve of Eratosthenes etc. about this topic I need help to understand how to reasoning with this kind of questions.

prime-numbers divisibility prime-factorization

share|cite|improve this question

asked Jan 17 at 11:44

PCNFPCNF

1338

$endgroup$

I'm trying to answer this question.

Are there positive integers $le200$ which have exactly 13 positive divisors? What about 14 divisors? If yes, write them. If no, explain why not.

Because I'm only familiar with prime factorization, the sieve of Eratosthenes etc. about this topic I need help to understand how to reasoning with this kind of questions.

prime-numbers divisibility prime-factorization

prime-numbers divisibility prime-factorization

share|cite|improve this question

asked Jan 17 at 11:44

PCNFPCNF

1338

share|cite|improve this question

asked Jan 17 at 11:44

PCNFPCNF

1338

share|cite|improve this question

share|cite|improve this question

asked Jan 17 at 11:44

PCNFPCNF

1338

asked Jan 17 at 11:44

PCNFPCNF

1338

asked Jan 17 at 11:44

PCNFPCNF

1338

1338

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you count repeated divisors? Like $2^3$ ?
  $endgroup$
  – Matti P.
  Jan 17 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Yes, I suppose
  $endgroup$
  – PCNF
  Jan 17 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I guess a number with 13 positive divisors can be written as the product $$ p_1 p_2 p_3 ldots p_{12}p_{13} $$ where $p_i$ are prime numbers, not necessarily different ones. So then the smallest of such number would be obtained by choosing $p_i=2$, resulting in $2^{13}>200$.
  $endgroup$
  – Matti P.
  Jan 17 at 12:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PCNF are you familiar with $tau(n)$ function ?
  $endgroup$
  – user626177
  Jan 17 at 12:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @someone only heard but never studied or used
  $endgroup$
  – PCNF
  Jan 17 at 14:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you count repeated divisors? Like $2^3$ ?
  $endgroup$
  – Matti P.
  Jan 17 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Yes, I suppose
  $endgroup$
  – PCNF
  Jan 17 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I guess a number with 13 positive divisors can be written as the product $$ p_1 p_2 p_3 ldots p_{12}p_{13} $$ where $p_i$ are prime numbers, not necessarily different ones. So then the smallest of such number would be obtained by choosing $p_i=2$, resulting in $2^{13}>200$.
  $endgroup$
  – Matti P.
  Jan 17 at 12:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PCNF are you familiar with $tau(n)$ function ?
  $endgroup$
  – user626177
  Jan 17 at 12:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @someone only heard but never studied or used
  $endgroup$
  – PCNF
  Jan 17 at 14:10
  

  
  
  
  

$begingroup$
Do you count repeated divisors? Like $2^3$ ?
$endgroup$
– Matti P.
Jan 17 at 12:26

$begingroup$
Do you count repeated divisors? Like $2^3$ ?
$endgroup$
– Matti P.
Jan 17 at 12:26

$begingroup$
@MattiP. Yes, I suppose
$endgroup$
– PCNF
Jan 17 at 12:29

$begingroup$
@MattiP. Yes, I suppose
$endgroup$
– PCNF
Jan 17 at 12:29

1

1

$begingroup$
Well, I guess a number with 13 positive divisors can be written as the product $$ p_1 p_2 p_3 ldots p_{12}p_{13} $$ where $p_i$ are prime numbers, not necessarily different ones. So then the smallest of such number would be obtained by choosing $p_i=2$, resulting in $2^{13}>200$.
$endgroup$
– Matti P.
Jan 17 at 12:32

$begingroup$
Well, I guess a number with 13 positive divisors can be written as the product $$ p_1 p_2 p_3 ldots p_{12}p_{13} $$ where $p_i$ are prime numbers, not necessarily different ones. So then the smallest of such number would be obtained by choosing $p_i=2$, resulting in $2^{13}>200$.
$endgroup$
– Matti P.
Jan 17 at 12:32

$begingroup$
@PCNF are you familiar with $tau(n)$ function ?
$endgroup$
– user626177
Jan 17 at 12:46

$begingroup$
@PCNF are you familiar with $tau(n)$ function ?
$endgroup$
– user626177
Jan 17 at 12:46

$begingroup$
@someone only heard but never studied or used
$endgroup$
– PCNF
Jan 17 at 14:10

$begingroup$
@someone only heard but never studied or used
$endgroup$
– PCNF
Jan 17 at 14:10

add a comment | 

1 Answer
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$begingroup$

For any number $n=p_1^{alpha_1}p_2^{alpha_2}p_3^{alpha_3}dots p_k^{alpha_k}$ the number of divisors (including $1$ and $n$) is $d=prod({alpha_i +1})$. If $d=13$, since $13$ is prime, there is only one possibility: $n=p_i^{12}$. The smallest such number is $2^{12}$ which is larger than $200$, so there are no numbers satisfying this requirement. If $d=14=2cdot 7$ candidate numbers will have the form $n=p_1cdot p_2^6$. The smallest such number is $3cdot 2^6 =192$ which meets your requirement of being less than $200$.

Added by edit: The formula $d=prod({alpha_i +1})$ is simply the $tau (n)$ referred to in the comments. Also, just for completeness sake, the $14$ divisors of $192$ are: $1,2,4,8,16,32,64,3,6,12,24,48,96,192$.

share|cite|improve this answer

edited Jan 17 at 15:59

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

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add a comment | 

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1 Answer
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1 Answer
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1

$begingroup$

For any number $n=p_1^{alpha_1}p_2^{alpha_2}p_3^{alpha_3}dots p_k^{alpha_k}$ the number of divisors (including $1$ and $n$) is $d=prod({alpha_i +1})$. If $d=13$, since $13$ is prime, there is only one possibility: $n=p_i^{12}$. The smallest such number is $2^{12}$ which is larger than $200$, so there are no numbers satisfying this requirement. If $d=14=2cdot 7$ candidate numbers will have the form $n=p_1cdot p_2^6$. The smallest such number is $3cdot 2^6 =192$ which meets your requirement of being less than $200$.

Added by edit: The formula $d=prod({alpha_i +1})$ is simply the $tau (n)$ referred to in the comments. Also, just for completeness sake, the $14$ divisors of $192$ are: $1,2,4,8,16,32,64,3,6,12,24,48,96,192$.

share|cite|improve this answer

edited Jan 17 at 15:59

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

$endgroup$

add a comment | 

1

$begingroup$

For any number $n=p_1^{alpha_1}p_2^{alpha_2}p_3^{alpha_3}dots p_k^{alpha_k}$ the number of divisors (including $1$ and $n$) is $d=prod({alpha_i +1})$. If $d=13$, since $13$ is prime, there is only one possibility: $n=p_i^{12}$. The smallest such number is $2^{12}$ which is larger than $200$, so there are no numbers satisfying this requirement. If $d=14=2cdot 7$ candidate numbers will have the form $n=p_1cdot p_2^6$. The smallest such number is $3cdot 2^6 =192$ which meets your requirement of being less than $200$.

Added by edit: The formula $d=prod({alpha_i +1})$ is simply the $tau (n)$ referred to in the comments. Also, just for completeness sake, the $14$ divisors of $192$ are: $1,2,4,8,16,32,64,3,6,12,24,48,96,192$.

share|cite|improve this answer

edited Jan 17 at 15:59

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

$endgroup$

add a comment | 

1

1

1

$begingroup$

For any number $n=p_1^{alpha_1}p_2^{alpha_2}p_3^{alpha_3}dots p_k^{alpha_k}$ the number of divisors (including $1$ and $n$) is $d=prod({alpha_i +1})$. If $d=13$, since $13$ is prime, there is only one possibility: $n=p_i^{12}$. The smallest such number is $2^{12}$ which is larger than $200$, so there are no numbers satisfying this requirement. If $d=14=2cdot 7$ candidate numbers will have the form $n=p_1cdot p_2^6$. The smallest such number is $3cdot 2^6 =192$ which meets your requirement of being less than $200$.

Added by edit: The formula $d=prod({alpha_i +1})$ is simply the $tau (n)$ referred to in the comments. Also, just for completeness sake, the $14$ divisors of $192$ are: $1,2,4,8,16,32,64,3,6,12,24,48,96,192$.

share|cite|improve this answer

edited Jan 17 at 15:59

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

$endgroup$

For any number $n=p_1^{alpha_1}p_2^{alpha_2}p_3^{alpha_3}dots p_k^{alpha_k}$ the number of divisors (including $1$ and $n$) is $d=prod({alpha_i +1})$. If $d=13$, since $13$ is prime, there is only one possibility: $n=p_i^{12}$. The smallest such number is $2^{12}$ which is larger than $200$, so there are no numbers satisfying this requirement. If $d=14=2cdot 7$ candidate numbers will have the form $n=p_1cdot p_2^6$. The smallest such number is $3cdot 2^6 =192$ which meets your requirement of being less than $200$.

Added by edit: The formula $d=prod({alpha_i +1})$ is simply the $tau (n)$ referred to in the comments. Also, just for completeness sake, the $14$ divisors of $192$ are: $1,2,4,8,16,32,64,3,6,12,24,48,96,192$.

share|cite|improve this answer

edited Jan 17 at 15:59

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

share|cite|improve this answer

share|cite|improve this answer

edited Jan 17 at 15:59

edited Jan 17 at 15:59

edited Jan 17 at 15:59

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

answered Jan 17 at 15:45

Keith BackmanKeith Backman

1,3681812

1,3681812

add a comment | 

add a comment | 

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