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How can I break up a larger number into smaller numbers and exponents?


Finding the maximum number with a certain Euler's totient valueEfficient way to compute $sum_{i=1}^n varphi(i) $Euler's totient function for large numbersDoes ($frac{p-1}{2})$! 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













0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $100=2^2cdot 5^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
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1 Answer
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active

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1












$begingroup$

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 $100=2^2cdot 5^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $100=2^2cdot 5^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $100=2^2cdot 5^2$.






share|cite|improve this answer









$endgroup$



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 $100=2^2cdot 5^2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 18 at 20:26









Dietrich BurdeDietrich Burde

81.5k648106




81.5k648106












  • $begingroup$
    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








  • 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


















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