Factorising $99999,00000,99999,00001$
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I am reading an article in The Mathematical Gazette about factorising the 20 digit number
$$N=99999,00000,99999,00001.$$
It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?
elementary-number-theory
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
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add a comment |
$begingroup$
I am reading an article in The Mathematical Gazette about factorising the 20 digit number
$$N=99999,00000,99999,00001.$$
It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?
elementary-number-theory
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
$endgroup$
add a comment |
$begingroup$
I am reading an article in The Mathematical Gazette about factorising the 20 digit number
$$N=99999,00000,99999,00001.$$
It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?
elementary-number-theory
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
$endgroup$
I am reading an article in The Mathematical Gazette about factorising the 20 digit number
$$N=99999,00000,99999,00001.$$
It is stated that $N=dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?
elementary-number-theory
elementary-number-theory
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
asked Dec 20 '18 at 10:31
rbirdrbird
1,20414
1,20414
add a comment |
add a comment |
1 Answer
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$begingroup$
If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).
Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
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1 Answer
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$begingroup$
If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).
Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
$endgroup$
add a comment |
$begingroup$
If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).
Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
$endgroup$
add a comment |
$begingroup$
If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).
Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
$endgroup$
If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}equiv -1 implies 10^{50}equiv 1pmod p,$$
i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}equiv 1pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}equiv 1pmod p$, but one can easily verify that $gcd(N,10^{10}-1)=1$).
Therefore, $d=50$. And Little Fermat says that $dmid(p-1)$.
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
answered Dec 20 '18 at 11:13
metamorphymetamorphy
3,6821621
3,6821621
add a comment |
add a comment |
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