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How can I break up a larger number into smaller numbers and exponents?
The Next CEO of Stack OverflowFinding the maximum number with a certain Euler's totient valueEfficient way to compute $sum_i=1^n varphi(i) $Euler's totient function for large numbersDoes ($fracp-12)$! satisty $x^2$ = 1modp?How many integers are divisible by at least one of the numbers $4$,$ 7$ and $33$?Inverse Euler totient functionInverse of Euler's phi (totient) functionHow is Carmichael's function subgroup of Euler's Totient function?Dirichlet ConvolutionSummation question with Euler's totient function
$begingroup$
I'm trying to compute the Euler's totient function for large numbers where I can't use the other rules in order to solve for the totient.
For example:
$theta(600) = (2^3 * 3 * 5^2)$
I am confused on how you can simplify 600 to to the numbers on the right hand of the expression.
number-theory totient-function
asked Mar 18 at 20:22
Code4lifeCode4life
61
$endgroup$
add a comment |
$begingroup$
I'm trying to compute the Euler's totient function for large numbers where I can't use the other rules in order to solve for the totient.
For example:
$theta(600) = (2^3 * 3 * 5^2)$
I am confused on how you can simplify 600 to to the numbers on the right hand of the expression.
number-theory totient-function
asked Mar 18 at 20:22
Code4lifeCode4life
61
$endgroup$
1
$begingroup$
You need to factorize $600$. To compute Euler's totient function there seems to be nothing simpler.
$endgroup$
– user647486
Mar 18 at 20:26
add a comment |
$begingroup$
I'm trying to compute the Euler's totient function for large numbers where I can't use the other rules in order to solve for the totient.
For example:
$theta(600) = (2^3 * 3 * 5^2)$
I am confused on how you can simplify 600 to to the numbers on the right hand of the expression.
number-theory totient-function
asked Mar 18 at 20:22
Code4lifeCode4life
61
$endgroup$
I'm trying to compute the Euler's totient function for large numbers where I can't use the other rules in order to solve for the totient.
For example:
$theta(600) = (2^3 * 3 * 5^2)$
I am confused on how you can simplify 600 to to the numbers on the right hand of the expression.
number-theory totient-function
number-theory totient-function
asked Mar 18 at 20:22
Code4lifeCode4life
61
asked Mar 18 at 20:22
Code4lifeCode4life
61
asked Mar 18 at 20:22
Code4lifeCode4life
61
asked Mar 18 at 20:22
Code4lifeCode4life
61
asked Mar 18 at 20:22
Code4lifeCode4life
61
61
1
$begingroup$
You need to factorize $600$. To compute Euler's totient function there seems to be nothing simpler.
$endgroup$
– user647486
Mar 18 at 20:26
add a comment |
1
$begingroup$
You need to factorize $600$. To compute Euler's totient function there seems to be nothing simpler.
$endgroup$
– user647486
Mar 18 at 20:26
1
1
$begingroup$
You need to factorize $600$. To compute Euler's totient function there seems to be nothing simpler.
$endgroup$
– user647486
Mar 18 at 20:26
$begingroup$
You need to factorize $600$. To compute Euler's totient function there seems to be nothing simpler.
$endgroup$
– user647486
Mar 18 at 20:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $600=2^3cdot 3cdot 5^2$ and
$$
phi(600)=phi(2^3)cdot phi(3)cdotphi(5^2)=4cdot 2cdot 20=160.
$$
The factorisation of $600$ is easy, since $100=2^2cdot 5^2$.
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
1
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
add a comment |
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1 Answer
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1 Answer
1
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$begingroup$
We have $600=2^3cdot 3cdot 5^2$ and
$$
phi(600)=phi(2^3)cdot phi(3)cdotphi(5^2)=4cdot 2cdot 20=160.
$$
The factorisation of $600$ is easy, since $100=2^2cdot 5^2$.
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
1
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
add a comment |
$begingroup$
We have $600=2^3cdot 3cdot 5^2$ and
$$
phi(600)=phi(2^3)cdot phi(3)cdotphi(5^2)=4cdot 2cdot 20=160.
$$
The factorisation of $600$ is easy, since $100=2^2cdot 5^2$.
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
1
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
add a comment |
$begingroup$
We have $600=2^3cdot 3cdot 5^2$ and
$$
phi(600)=phi(2^3)cdot phi(3)cdotphi(5^2)=4cdot 2cdot 20=160.
$$
The factorisation of $600$ is easy, since $100=2^2cdot 5^2$.
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
We have $600=2^3cdot 3cdot 5^2$ and
$$
phi(600)=phi(2^3)cdot phi(3)cdotphi(5^2)=4cdot 2cdot 20=160.
$$
The factorisation of $600$ is easy, since $100=2^2cdot 5^2$.
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
answered Mar 18 at 20:26
Dietrich BurdeDietrich Burde
81.6k648106
81.6k648106
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
1
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
add a comment |
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
1
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
$begingroup$
Okay that makes sense, but when I tried factoring $100 = 10^2$ and solved for the Euler's totient I got 180. $theta(600) = 10^2 * 6 = (10^2 - 10) * 2 = 180$. The Euler totient of 6 is 2.
$endgroup$
– Code4life
Mar 18 at 20:36
1
1
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
But $10$ is not prime, so $phi(10^2)neq10^2-10$.
$endgroup$
– Servaes
Mar 18 at 20:40
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
$begingroup$
Ohhh okay base numbers have to prime that makes sense
$endgroup$
– Code4life
Mar 18 at 20:55
add a comment |
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1
$begingroup$
You need to factorize $600$. To compute Euler's totient function there seems to be nothing simpler.
$endgroup$
– user647486
Mar 18 at 20:26