How many positive integers $le 1260$ are relatively prime to $1260$? [duplicate]
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Find all integers less than $m$ that are relatively prime to it
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I have no idea how to solve this problem.
Is there a general formula to compute the quantity of such numbers?
number-theory coprime
edited Jan 1 at 10:05
Key Flex
8,31261233
asked Jan 1 at 9:49
MelinaMelina
83
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Jan 1 at 10:19
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.
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$begingroup$
This question already has an answer here:
Find all integers less than $m$ that are relatively prime to it
2 answers
I have no idea how to solve this problem.
Is there a general formula to compute the quantity of such numbers?
number-theory coprime
edited Jan 1 at 10:05
Key Flex
8,31261233
asked Jan 1 at 9:49
MelinaMelina
83
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marked as duplicate by Dietrich Burde number-theory
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Jan 1 at 10:19
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.
2
$begingroup$
en.wikipedia.org/wiki/Euler%27s_totient_function
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– Lord Shark the Unknown
Jan 1 at 9:50
add a comment |
$begingroup$
This question already has an answer here:
Find all integers less than $m$ that are relatively prime to it
2 answers
I have no idea how to solve this problem.
Is there a general formula to compute the quantity of such numbers?
number-theory coprime
edited Jan 1 at 10:05
Key Flex
8,31261233
asked Jan 1 at 9:49
MelinaMelina
83
$endgroup$
This question already has an answer here:
Find all integers less than $m$ that are relatively prime to it
2 answers
I have no idea how to solve this problem.
Is there a general formula to compute the quantity of such numbers?
This question already has an answer here:
Find all integers less than $m$ that are relatively prime to it
2 answers
number-theory coprime
number-theory coprime
edited Jan 1 at 10:05
Key Flex
8,31261233
asked Jan 1 at 9:49
MelinaMelina
83
edited Jan 1 at 10:05
Key Flex
8,31261233
asked Jan 1 at 9:49
MelinaMelina
83
edited Jan 1 at 10:05
Key Flex
8,31261233
edited Jan 1 at 10:05
Key Flex
8,31261233
edited Jan 1 at 10:05
Key Flex
8,31261233
8,31261233
asked Jan 1 at 9:49
MelinaMelina
83
asked Jan 1 at 9:49
MelinaMelina
83
asked Jan 1 at 9:49
MelinaMelina
83
83
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Jan 1 at 10:19
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Jan 1 at 10:19
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.
2
$begingroup$
en.wikipedia.org/wiki/Euler%27s_totient_function
$endgroup$
– Lord Shark the Unknown
Jan 1 at 9:50
add a comment |
2
$begingroup$
en.wikipedia.org/wiki/Euler%27s_totient_function
$endgroup$
– Lord Shark the Unknown
Jan 1 at 9:50
2
2
$begingroup$
en.wikipedia.org/wiki/Euler%27s_totient_function
$endgroup$
– Lord Shark the Unknown
Jan 1 at 9:50
$begingroup$
en.wikipedia.org/wiki/Euler%27s_totient_function
$endgroup$
– Lord Shark the Unknown
Jan 1 at 9:50
add a comment |
3 Answers
3
active
oldest
votes
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As the comment stated, you want to use the Euler's totient function.
A good way to calculate the totient function is the use of the product formula. We first note that the prime factorisation of $1260 = 2^2 cdot 3^2 cdot 5 cdot 7$.
We then calculate $$phi(1260) = 1260 cdot (1 - frac{1}{2})(1 - frac{1}{3})(1 - frac{1}{5})(1 - frac{1}{7}) = 288$$.
answered Jan 1 at 10:02
twnlytwnly
1,0631213
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add a comment |
$begingroup$
As Lord Shark the Unknown pointed out, the number of positive integers less than or equal to a given number that are coprime to it is called the Euler Totient Function. The easiest way to calculate this is
$$varphi (n)=nprod _{pmid n}left(1-{frac {1}{p}}right)$$
where the product is over all prime numbers p that divide n. So for 1260, the prime factors are 2,3,5,7 so we have
$$varphi (1260)=1260 (1-frac12)(1-frac13)(1-frac15)(1-frac17) = 288$$
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
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add a comment |
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It works like this.
The primes that divide 1260 are 2, 3, 5, 7.
For each of these, there is only one remainder that indicates the number is a multiple of the prime: that is 0.
So the number of co-primes is thus all combinations of non-division, ie
[ (2-1)(3-1)(5-1)(7-1)/2.3.5.7 ] * 1260.
We see the square brackets give 48/210, and this by 1260 gives 288.
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As the comment stated, you want to use the Euler's totient function.
A good way to calculate the totient function is the use of the product formula. We first note that the prime factorisation of $1260 = 2^2 cdot 3^2 cdot 5 cdot 7$.
We then calculate $$phi(1260) = 1260 cdot (1 - frac{1}{2})(1 - frac{1}{3})(1 - frac{1}{5})(1 - frac{1}{7}) = 288$$.
answered Jan 1 at 10:02
twnlytwnly
1,0631213
$endgroup$
add a comment |
$begingroup$
As the comment stated, you want to use the Euler's totient function.
A good way to calculate the totient function is the use of the product formula. We first note that the prime factorisation of $1260 = 2^2 cdot 3^2 cdot 5 cdot 7$.
We then calculate $$phi(1260) = 1260 cdot (1 - frac{1}{2})(1 - frac{1}{3})(1 - frac{1}{5})(1 - frac{1}{7}) = 288$$.
answered Jan 1 at 10:02
twnlytwnly
1,0631213
$endgroup$
add a comment |
$begingroup$
As the comment stated, you want to use the Euler's totient function.
A good way to calculate the totient function is the use of the product formula. We first note that the prime factorisation of $1260 = 2^2 cdot 3^2 cdot 5 cdot 7$.
We then calculate $$phi(1260) = 1260 cdot (1 - frac{1}{2})(1 - frac{1}{3})(1 - frac{1}{5})(1 - frac{1}{7}) = 288$$.
answered Jan 1 at 10:02
twnlytwnly
1,0631213
$endgroup$
As the comment stated, you want to use the Euler's totient function.
A good way to calculate the totient function is the use of the product formula. We first note that the prime factorisation of $1260 = 2^2 cdot 3^2 cdot 5 cdot 7$.
We then calculate $$phi(1260) = 1260 cdot (1 - frac{1}{2})(1 - frac{1}{3})(1 - frac{1}{5})(1 - frac{1}{7}) = 288$$.
answered Jan 1 at 10:02
twnlytwnly
1,0631213
answered Jan 1 at 10:02
twnlytwnly
1,0631213
answered Jan 1 at 10:02
twnlytwnly
1,0631213
answered Jan 1 at 10:02
twnlytwnly
1,0631213
1,0631213
add a comment |
add a comment |
$begingroup$
As Lord Shark the Unknown pointed out, the number of positive integers less than or equal to a given number that are coprime to it is called the Euler Totient Function. The easiest way to calculate this is
$$varphi (n)=nprod _{pmid n}left(1-{frac {1}{p}}right)$$
where the product is over all prime numbers p that divide n. So for 1260, the prime factors are 2,3,5,7 so we have
$$varphi (1260)=1260 (1-frac12)(1-frac13)(1-frac15)(1-frac17) = 288$$
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
add a comment |
$begingroup$
As Lord Shark the Unknown pointed out, the number of positive integers less than or equal to a given number that are coprime to it is called the Euler Totient Function. The easiest way to calculate this is
$$varphi (n)=nprod _{pmid n}left(1-{frac {1}{p}}right)$$
where the product is over all prime numbers p that divide n. So for 1260, the prime factors are 2,3,5,7 so we have
$$varphi (1260)=1260 (1-frac12)(1-frac13)(1-frac15)(1-frac17) = 288$$
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
add a comment |
$begingroup$
As Lord Shark the Unknown pointed out, the number of positive integers less than or equal to a given number that are coprime to it is called the Euler Totient Function. The easiest way to calculate this is
$$varphi (n)=nprod _{pmid n}left(1-{frac {1}{p}}right)$$
where the product is over all prime numbers p that divide n. So for 1260, the prime factors are 2,3,5,7 so we have
$$varphi (1260)=1260 (1-frac12)(1-frac13)(1-frac15)(1-frac17) = 288$$
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
As Lord Shark the Unknown pointed out, the number of positive integers less than or equal to a given number that are coprime to it is called the Euler Totient Function. The easiest way to calculate this is
$$varphi (n)=nprod _{pmid n}left(1-{frac {1}{p}}right)$$
where the product is over all prime numbers p that divide n. So for 1260, the prime factors are 2,3,5,7 so we have
$$varphi (1260)=1260 (1-frac12)(1-frac13)(1-frac15)(1-frac17) = 288$$
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
answered Jan 1 at 10:02
Erik ParkinsonErik Parkinson
1,16519
1,16519
add a comment |
add a comment |
$begingroup$
It works like this.
The primes that divide 1260 are 2, 3, 5, 7.
For each of these, there is only one remainder that indicates the number is a multiple of the prime: that is 0.
So the number of co-primes is thus all combinations of non-division, ie
[ (2-1)(3-1)(5-1)(7-1)/2.3.5.7 ] * 1260.
We see the square brackets give 48/210, and this by 1260 gives 288.
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
$endgroup$
add a comment |
$begingroup$
It works like this.
The primes that divide 1260 are 2, 3, 5, 7.
For each of these, there is only one remainder that indicates the number is a multiple of the prime: that is 0.
So the number of co-primes is thus all combinations of non-division, ie
[ (2-1)(3-1)(5-1)(7-1)/2.3.5.7 ] * 1260.
We see the square brackets give 48/210, and this by 1260 gives 288.
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
$endgroup$
add a comment |
$begingroup$
It works like this.
The primes that divide 1260 are 2, 3, 5, 7.
For each of these, there is only one remainder that indicates the number is a multiple of the prime: that is 0.
So the number of co-primes is thus all combinations of non-division, ie
[ (2-1)(3-1)(5-1)(7-1)/2.3.5.7 ] * 1260.
We see the square brackets give 48/210, and this by 1260 gives 288.
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
$endgroup$
It works like this.
The primes that divide 1260 are 2, 3, 5, 7.
For each of these, there is only one remainder that indicates the number is a multiple of the prime: that is 0.
So the number of co-primes is thus all combinations of non-division, ie
[ (2-1)(3-1)(5-1)(7-1)/2.3.5.7 ] * 1260.
We see the square brackets give 48/210, and this by 1260 gives 288.
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
answered Jan 1 at 10:12
wendy.kriegerwendy.krieger
5,84711426
5,84711426
add a comment |
add a comment |
2
$begingroup$
en.wikipedia.org/wiki/Euler%27s_totient_function
$endgroup$
– Lord Shark the Unknown
Jan 1 at 9:50