$c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$
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0
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How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?
I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.
Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$
$a^{p-1} equiv 1 pmod p$
And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.
That still doesn't help me though...
modular-arithmetic
asked Nov 22 at 10:49
Math Dummy
276
add a comment |
up vote
0
down vote
favorite
How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?
I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.
Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$
$a^{p-1} equiv 1 pmod p$
And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.
That still doesn't help me though...
modular-arithmetic
asked Nov 22 at 10:49
Math Dummy
276
$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52
My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?
I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.
Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$
$a^{p-1} equiv 1 pmod p$
And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.
That still doesn't help me though...
modular-arithmetic
asked Nov 22 at 10:49
Math Dummy
276
How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?
I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.
Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$
$a^{p-1} equiv 1 pmod p$
And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.
That still doesn't help me though...
modular-arithmetic
modular-arithmetic
asked Nov 22 at 10:49
Math Dummy
276
asked Nov 22 at 10:49
Math Dummy
276
asked Nov 22 at 10:49
Math Dummy
276
asked Nov 22 at 10:49
Math Dummy
276
asked Nov 22 at 10:49
Math Dummy
276
276
$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52
My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56
add a comment |
$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52
My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56
$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52
$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52
My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56
My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56
add a comment |
2 Answers
2
active
oldest
votes
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0
down vote
accepted
Fermat's little theorem provides a solution:
$$
c cdot 11 equiv 23 bmod 103
implies
c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
$$
Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...
answered Nov 22 at 11:10
lhf
162k9166385
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up vote
0
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By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Fermat's little theorem provides a solution:
$$
c cdot 11 equiv 23 bmod 103
implies
c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
$$
Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...
answered Nov 22 at 11:10
lhf
162k9166385
add a comment |
up vote
0
down vote
accepted
Fermat's little theorem provides a solution:
$$
c cdot 11 equiv 23 bmod 103
implies
c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
$$
Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...
answered Nov 22 at 11:10
lhf
162k9166385
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Fermat's little theorem provides a solution:
$$
c cdot 11 equiv 23 bmod 103
implies
c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
$$
Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...
answered Nov 22 at 11:10
lhf
162k9166385
Fermat's little theorem provides a solution:
$$
c cdot 11 equiv 23 bmod 103
implies
c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
$$
Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...
answered Nov 22 at 11:10
lhf
162k9166385
answered Nov 22 at 11:10
lhf
162k9166385
answered Nov 22 at 11:10
lhf
162k9166385
answered Nov 22 at 11:10
lhf
162k9166385
162k9166385
add a comment |
add a comment |
up vote
0
down vote
By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
add a comment |
up vote
0
down vote
By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
add a comment |
up vote
0
down vote
up vote
0
down vote
By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
answered Nov 22 at 19:50
Bill Dubuque
207k29189625
207k29189625
add a comment |
add a comment |
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$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52
My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56