Modulo calculation
$begingroup$
I am stuck on this modulo calculation:
$$718^{143} pmod{1260}$$
I have tried using the Euler totient function with no success ($phi(1260) = 288)$. I think I could solve it using the Chinese remainder theorem but I am guessing there is a faster way.
Any help is appreciated, thanks in advance!
discrete-mathematics modular-arithmetic
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
$endgroup$
add a comment |
$begingroup$
I am stuck on this modulo calculation:
$$718^{143} pmod{1260}$$
I have tried using the Euler totient function with no success ($phi(1260) = 288)$. I think I could solve it using the Chinese remainder theorem but I am guessing there is a faster way.
Any help is appreciated, thanks in advance!
discrete-mathematics modular-arithmetic
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
$endgroup$
$begingroup$
I expect the most standard way to do this is exponentiation by squaring. en.wikipedia.org/wiki/Exponentiation_by_squaring But there might be a neat trick in this particular case.
$endgroup$
– SmileyCraft
Jan 22 at 18:28
1
$begingroup$
Using Carmichael function might be very helpful... For your case, this states that $$a^{lambda (1260)}=a^{12}equiv 1 mod 1260 ;; forall a, nin mathbb N : text{gcd}(a,n)=1$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:27
add a comment |
$begingroup$
I am stuck on this modulo calculation:
$$718^{143} pmod{1260}$$
I have tried using the Euler totient function with no success ($phi(1260) = 288)$. I think I could solve it using the Chinese remainder theorem but I am guessing there is a faster way.
Any help is appreciated, thanks in advance!
discrete-mathematics modular-arithmetic
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
$endgroup$
I am stuck on this modulo calculation:
$$718^{143} pmod{1260}$$
I have tried using the Euler totient function with no success ($phi(1260) = 288)$. I think I could solve it using the Chinese remainder theorem but I am guessing there is a faster way.
Any help is appreciated, thanks in advance!
discrete-mathematics modular-arithmetic
discrete-mathematics modular-arithmetic
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
asked Jan 22 at 18:25
Algebruh ghostAlgebruh ghost
252
252
$begingroup$
I expect the most standard way to do this is exponentiation by squaring. en.wikipedia.org/wiki/Exponentiation_by_squaring But there might be a neat trick in this particular case.
$endgroup$
– SmileyCraft
Jan 22 at 18:28
1
$begingroup$
Using Carmichael function might be very helpful... For your case, this states that $$a^{lambda (1260)}=a^{12}equiv 1 mod 1260 ;; forall a, nin mathbb N : text{gcd}(a,n)=1$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:27
add a comment |
$begingroup$
I expect the most standard way to do this is exponentiation by squaring. en.wikipedia.org/wiki/Exponentiation_by_squaring But there might be a neat trick in this particular case.
$endgroup$
– SmileyCraft
Jan 22 at 18:28
1
$begingroup$
Using Carmichael function might be very helpful... For your case, this states that $$a^{lambda (1260)}=a^{12}equiv 1 mod 1260 ;; forall a, nin mathbb N : text{gcd}(a,n)=1$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:27
$begingroup$
I expect the most standard way to do this is exponentiation by squaring. en.wikipedia.org/wiki/Exponentiation_by_squaring But there might be a neat trick in this particular case.
$endgroup$
– SmileyCraft
Jan 22 at 18:28
$begingroup$
I expect the most standard way to do this is exponentiation by squaring. en.wikipedia.org/wiki/Exponentiation_by_squaring But there might be a neat trick in this particular case.
$endgroup$
– SmileyCraft
Jan 22 at 18:28
1
1
$begingroup$
Using Carmichael function might be very helpful... For your case, this states that $$a^{lambda (1260)}=a^{12}equiv 1 mod 1260 ;; forall a, nin mathbb N : text{gcd}(a,n)=1$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:27
$begingroup$
Using Carmichael function might be very helpful... For your case, this states that $$a^{lambda (1260)}=a^{12}equiv 1 mod 1260 ;; forall a, nin mathbb N : text{gcd}(a,n)=1$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints: $718=2cdot 359$, $ 1260=2cdot 630$, $ varphi(630)=144$,
So, $359^{143}equiv 359^{-1}pmod{630}$ which can be found using the Euclidean algorithm, say it's $x$.
We also need to calculate $2^{142}pmod{630}$, say it's $y$, and then we can combine them to
$$718^{143}=2cdot 2^{142}cdot 359^{143}equiv 2xypmod{2cdot 630}$$
using that $aequiv bpmod mimplies 2aequiv 2bpmod{2m}$.
answered Jan 22 at 19:15
BerciBerci
61.2k23674
$endgroup$
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Hints: $718=2cdot 359$, $ 1260=2cdot 630$, $ varphi(630)=144$,
So, $359^{143}equiv 359^{-1}pmod{630}$ which can be found using the Euclidean algorithm, say it's $x$.
We also need to calculate $2^{142}pmod{630}$, say it's $y$, and then we can combine them to
$$718^{143}=2cdot 2^{142}cdot 359^{143}equiv 2xypmod{2cdot 630}$$
using that $aequiv bpmod mimplies 2aequiv 2bpmod{2m}$.
answered Jan 22 at 19:15
BerciBerci
61.2k23674
$endgroup$
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
add a comment |
$begingroup$
Hints: $718=2cdot 359$, $ 1260=2cdot 630$, $ varphi(630)=144$,
So, $359^{143}equiv 359^{-1}pmod{630}$ which can be found using the Euclidean algorithm, say it's $x$.
We also need to calculate $2^{142}pmod{630}$, say it's $y$, and then we can combine them to
$$718^{143}=2cdot 2^{142}cdot 359^{143}equiv 2xypmod{2cdot 630}$$
using that $aequiv bpmod mimplies 2aequiv 2bpmod{2m}$.
answered Jan 22 at 19:15
BerciBerci
61.2k23674
$endgroup$
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
add a comment |
$begingroup$
Hints: $718=2cdot 359$, $ 1260=2cdot 630$, $ varphi(630)=144$,
So, $359^{143}equiv 359^{-1}pmod{630}$ which can be found using the Euclidean algorithm, say it's $x$.
We also need to calculate $2^{142}pmod{630}$, say it's $y$, and then we can combine them to
$$718^{143}=2cdot 2^{142}cdot 359^{143}equiv 2xypmod{2cdot 630}$$
using that $aequiv bpmod mimplies 2aequiv 2bpmod{2m}$.
answered Jan 22 at 19:15
BerciBerci
61.2k23674
$endgroup$
Hints: $718=2cdot 359$, $ 1260=2cdot 630$, $ varphi(630)=144$,
So, $359^{143}equiv 359^{-1}pmod{630}$ which can be found using the Euclidean algorithm, say it's $x$.
We also need to calculate $2^{142}pmod{630}$, say it's $y$, and then we can combine them to
$$718^{143}=2cdot 2^{142}cdot 359^{143}equiv 2xypmod{2cdot 630}$$
using that $aequiv bpmod mimplies 2aequiv 2bpmod{2m}$.
answered Jan 22 at 19:15
BerciBerci
61.2k23674
answered Jan 22 at 19:15
BerciBerci
61.2k23674
answered Jan 22 at 19:15
BerciBerci
61.2k23674
answered Jan 22 at 19:15
BerciBerci
61.2k23674
61.2k23674
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
add a comment |
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
$begingroup$
Why do people feel completely justified in using a calculator to multiply two large numbers but still want to do this Mod problems by hand? It is straightforward on a computer: $1192$. Done.
$endgroup$
– David G. Stork
Jan 22 at 19:20
add a comment |
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$begingroup$
I expect the most standard way to do this is exponentiation by squaring. en.wikipedia.org/wiki/Exponentiation_by_squaring But there might be a neat trick in this particular case.
$endgroup$
– SmileyCraft
Jan 22 at 18:28
1
$begingroup$
Using Carmichael function might be very helpful... For your case, this states that $$a^{lambda (1260)}=a^{12}equiv 1 mod 1260 ;; forall a, nin mathbb N : text{gcd}(a,n)=1$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:27