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How many integers are there that are not divisible by any prime larger than 20 and not divisible by the square of any prime?

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6

$begingroup$

I tackled the problem in the following way but i'm not sure if i'm correct.

I need the count of the numbers that have in their prime factorization only primes p such that $p lt 20$ and those numbers can't be more than once in the prime factorization (right?)

So, the amount of numbers that can be expressed this way are all the subsets of the set ${2,3,5,7,11,13,17,19} = 2^8$.

Correct me if I'm wrong.

number-theory prime-numbers divisibility prime-factorization

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edited 8 hours ago

David G. Stork

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Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

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• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If the question was asking about natural numbers, you would be correct.
  $endgroup$
  – Don Thousand
  8 hours ago
  

  
  
  
  

add a comment
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6

$begingroup$

I tackled the problem in the following way but i'm not sure if i'm correct.

I need the count of the numbers that have in their prime factorization only primes p such that $p lt 20$ and those numbers can't be more than once in the prime factorization (right?)

So, the amount of numbers that can be expressed this way are all the subsets of the set ${2,3,5,7,11,13,17,19} = 2^8$.

Correct me if I'm wrong.

number-theory prime-numbers divisibility prime-factorization

share|cite|improve this question

edited 8 hours ago

David G. Stork

15.6k44 gold badges2020 silver badges4040 bronze badges

asked 8 hours ago

Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

2922 bronze badges

New contributor

Andrés Felipe Vargas Fontecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If the question was asking about natural numbers, you would be correct.
  $endgroup$
  – Don Thousand
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I tackled the problem in the following way but i'm not sure if i'm correct.

I need the count of the numbers that have in their prime factorization only primes p such that $p lt 20$ and those numbers can't be more than once in the prime factorization (right?)

So, the amount of numbers that can be expressed this way are all the subsets of the set ${2,3,5,7,11,13,17,19} = 2^8$.

Correct me if I'm wrong.

number-theory prime-numbers divisibility prime-factorization

share|cite|improve this question

edited 8 hours ago

David G. Stork

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asked 8 hours ago

Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

2922 bronze badges

New contributor

Andrés Felipe Vargas Fontecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

David G. Stork

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asked 8 hours ago

Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

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share|cite|improve this question

edited 8 hours ago

David G. Stork

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asked 8 hours ago

Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

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edited 8 hours ago

David G. Stork

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edited 8 hours ago

David G. Stork

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asked 8 hours ago

Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

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asked 8 hours ago

Andrés Felipe Vargas FontechaAndrés Felipe Vargas Fontecha

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4

$begingroup$

Your method is perfectly correct, but if it is actually about integers then you need to include negatives as well to have twice as many.

I suspect this is not the case however since then if you said "primes $p<20$" you'd probably have to include negative primes as well, in which case there would be infinitely many.

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edited 8 hours ago

answered 8 hours ago

Matt SamuelMatt Samuel

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4

$begingroup$

Your method is perfectly correct, but if it is actually about integers then you need to include negatives as well to have twice as many.

I suspect this is not the case however since then if you said "primes $p<20$" you'd probably have to include negative primes as well, in which case there would be infinitely many.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Matt SamuelMatt Samuel

42.6k66 gold badges4242 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Your method is perfectly correct, but if it is actually about integers then you need to include negatives as well to have twice as many.

I suspect this is not the case however since then if you said "primes $p<20$" you'd probably have to include negative primes as well, in which case there would be infinitely many.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Matt SamuelMatt Samuel

42.6k66 gold badges4242 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Your method is perfectly correct, but if it is actually about integers then you need to include negatives as well to have twice as many.

I suspect this is not the case however since then if you said "primes $p<20$" you'd probably have to include negative primes as well, in which case there would be infinitely many.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Matt SamuelMatt Samuel

42.6k66 gold badges4242 silver badges7474 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Matt SamuelMatt Samuel

42.6k66 gold badges4242 silver badges7474 bronze badges

answered 8 hours ago

Matt SamuelMatt Samuel

42.6k66 gold badges4242 silver badges7474 bronze badges

answered 8 hours ago

Matt SamuelMatt Samuel

42.6k66 gold badges4242 silver badges7474 bronze badges