Mixing
Calculate: $x=2^{12} pmod{13}$ in $mathbb{Z}_{13}$
$begingroup$
Calculate: $x=2^{12} pmod{13}$ in $mathbb{Z}_{13}$ by using Fermat's Little Theorem.
So I tried it like this but I don't know if I did it correctly?
Since the $text{gcd}(2,13)=1$ we can use Fermat which says that
$$2^{12}equiv1pmod{13}$$
So we can write:
$$x equiv 2^{12}equiv1pmod{13}$$
Thus the solution will be $1 pmod{13}$ ?
elementary-number-theory proof-verification modular-arithmetic
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
$endgroup$
|
show 1 more comment
$begingroup$
Calculate: $x=2^{12} pmod{13}$ in $mathbb{Z}_{13}$ by using Fermat's Little Theorem.
So I tried it like this but I don't know if I did it correctly?
Since the $text{gcd}(2,13)=1$ we can use Fermat which says that
$$2^{12}equiv1pmod{13}$$
So we can write:
$$x equiv 2^{12}equiv1pmod{13}$$
Thus the solution will be $1 pmod{13}$ ?
elementary-number-theory proof-verification modular-arithmetic
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
$endgroup$
4
$begingroup$
You're not using Fermat correctly (well you are, but not for the reason you think you are) : the reason you can use it is because 13 is prime, and because 13 does not divide 2, not only because $gcd(2,13)=1$
$endgroup$
– Astyx
Nov 17 '16 at 17:51
1
$begingroup$
@Astyx If $p$ is prime, $ainmathbb Z^+$, then $pnmid a$ is equivalent to $gcd(p,a)=1$.
$endgroup$
– user236182
Nov 17 '16 at 17:52
2
$begingroup$
@user236182 I absolutely agree, hence the "not only because ..." :)
$endgroup$
– Astyx
Nov 17 '16 at 17:55
$begingroup$
Thank you very much for these info! Is the rest really fine how I wrote it? It seems like a strange theorem to me and the way it's written. Really correct?
$endgroup$
– cnmesr
Nov 17 '16 at 17:57
$begingroup$
It's correct.${}$
$endgroup$
– user236182
Nov 17 '16 at 17:59
|
show 1 more comment
$begingroup$
Calculate: $x=2^{12} pmod{13}$ in $mathbb{Z}_{13}$ by using Fermat's Little Theorem.
So I tried it like this but I don't know if I did it correctly?
Since the $text{gcd}(2,13)=1$ we can use Fermat which says that
$$2^{12}equiv1pmod{13}$$
So we can write:
$$x equiv 2^{12}equiv1pmod{13}$$
Thus the solution will be $1 pmod{13}$ ?
elementary-number-theory proof-verification modular-arithmetic
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
$endgroup$
Calculate: $x=2^{12} pmod{13}$ in $mathbb{Z}_{13}$ by using Fermat's Little Theorem.
So I tried it like this but I don't know if I did it correctly?
Since the $text{gcd}(2,13)=1$ we can use Fermat which says that
$$2^{12}equiv1pmod{13}$$
So we can write:
$$x equiv 2^{12}equiv1pmod{13}$$
Thus the solution will be $1 pmod{13}$ ?
elementary-number-theory proof-verification modular-arithmetic
elementary-number-theory proof-verification modular-arithmetic
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
edited Jan 4 at 1:15
Eevee Trainer
5,4791936
5,4791936
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
asked Nov 17 '16 at 17:44
cnmesrcnmesr
2,16341942
2,16341942
4
$begingroup$
You're not using Fermat correctly (well you are, but not for the reason you think you are) : the reason you can use it is because 13 is prime, and because 13 does not divide 2, not only because $gcd(2,13)=1$
$endgroup$
– Astyx
Nov 17 '16 at 17:51
1
$begingroup$
@Astyx If $p$ is prime, $ainmathbb Z^+$, then $pnmid a$ is equivalent to $gcd(p,a)=1$.
$endgroup$
– user236182
Nov 17 '16 at 17:52
2
$begingroup$
@user236182 I absolutely agree, hence the "not only because ..." :)
$endgroup$
– Astyx
Nov 17 '16 at 17:55
$begingroup$
Thank you very much for these info! Is the rest really fine how I wrote it? It seems like a strange theorem to me and the way it's written. Really correct?
$endgroup$
– cnmesr
Nov 17 '16 at 17:57
$begingroup$
It's correct.${}$
$endgroup$
– user236182
Nov 17 '16 at 17:59
|
show 1 more comment
4
$begingroup$
You're not using Fermat correctly (well you are, but not for the reason you think you are) : the reason you can use it is because 13 is prime, and because 13 does not divide 2, not only because $gcd(2,13)=1$
$endgroup$
– Astyx
Nov 17 '16 at 17:51
1
$begingroup$
@Astyx If $p$ is prime, $ainmathbb Z^+$, then $pnmid a$ is equivalent to $gcd(p,a)=1$.
$endgroup$
– user236182
Nov 17 '16 at 17:52
2
$begingroup$
@user236182 I absolutely agree, hence the "not only because ..." :)
$endgroup$
– Astyx
Nov 17 '16 at 17:55
$begingroup$
Thank you very much for these info! Is the rest really fine how I wrote it? It seems like a strange theorem to me and the way it's written. Really correct?
$endgroup$
– cnmesr
Nov 17 '16 at 17:57
$begingroup$
It's correct.${}$
$endgroup$
– user236182
Nov 17 '16 at 17:59
4
4
$begingroup$
You're not using Fermat correctly (well you are, but not for the reason you think you are) : the reason you can use it is because 13 is prime, and because 13 does not divide 2, not only because $gcd(2,13)=1$
$endgroup$
– Astyx
Nov 17 '16 at 17:51
$begingroup$
You're not using Fermat correctly (well you are, but not for the reason you think you are) : the reason you can use it is because 13 is prime, and because 13 does not divide 2, not only because $gcd(2,13)=1$
$endgroup$
– Astyx
Nov 17 '16 at 17:51
1
1
$begingroup$
@Astyx If $p$ is prime, $ainmathbb Z^+$, then $pnmid a$ is equivalent to $gcd(p,a)=1$.
$endgroup$
– user236182
Nov 17 '16 at 17:52
$begingroup$
@Astyx If $p$ is prime, $ainmathbb Z^+$, then $pnmid a$ is equivalent to $gcd(p,a)=1$.
$endgroup$
– user236182
Nov 17 '16 at 17:52
2
2
$begingroup$
@user236182 I absolutely agree, hence the "not only because ..." :)
$endgroup$
– Astyx
Nov 17 '16 at 17:55
$begingroup$
@user236182 I absolutely agree, hence the "not only because ..." :)
$endgroup$
– Astyx
Nov 17 '16 at 17:55
$begingroup$
Thank you very much for these info! Is the rest really fine how I wrote it? It seems like a strange theorem to me and the way it's written. Really correct?
$endgroup$
– cnmesr
Nov 17 '16 at 17:57
$begingroup$
Thank you very much for these info! Is the rest really fine how I wrote it? It seems like a strange theorem to me and the way it's written. Really correct?
$endgroup$
– cnmesr
Nov 17 '16 at 17:57
$begingroup$
It's correct.${}$
$endgroup$
– user236182
Nov 17 '16 at 17:59
$begingroup$
It's correct.${}$
$endgroup$
– user236182
Nov 17 '16 at 17:59
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
It might worth noting that your solution is easy to check in this case: it is worth memorizing your powers of $2$, or at least certain specific ones (e.g. $2^{10}$). In this case, $2^{12} = 4,096$, which leaves a remainder of $1$ when divided by $13$. In this case, you thus use that to verify you're correct, but it's understandable that this is (a) not the method intended to be used and (b) might not be feasible for the general case.
Anyhow, let us remember Fermat's Little Theorem:
Fermat's Little Theorem: Let $p$ be prime, and $a$ an integer. Then $a^p - a$ is a multiple of $p$. This can be stated symbolically in a few ways: namely:
$$p | (a^p - a) ;;;;;;;;;; frac{a^p - a}{p} = k in mathbb{Z} ;;;;;;;;;; a^p equiv a pmod p$$
Corollary/Implication: If $p not | a$, then this implies $a^{p-1} - 1$ is a multiple of $p$. Again, equivalent formulations:
$$p | (a^{p-1} - 1) ;;;;;;;;;; frac{a^{p-1} - 1}{p} = k in mathbb{Z} ;;;;;;;;;; a^{p-1} equiv 1 pmod p$$
We use this latter form of Fermat's Little Theorem in this problem, with $a = 2, p = 13$.
Then by Fermat's Little Theorem, since $p = 13$ is a prime which does divide $a = 2$, we can claim the following:
$$2^{13-1} equiv 1 pmod {13} iff 2^{12} equiv 1 pmod {13}$$
When $mathbb{Z}_{13}$ denotes the integers mod $13$, then it is clear $1 in mathbb{Z}_{13}$, and thus $x = 1$ in this problem.
Your proof is more or less right, but Fermat's Little Theorem is not used correctly. Having $gcd(2,13) = 1$ is not a sufficient condition for the corollary of Fermat's Little Theorem: one of them must also be prime. (Then $gcd(a,p) = 1$ would imply the necessary condition of non-divisibility.) As an example, $gcd(15, 14) = 1$ but neither are prime, which means Fermat's Little Theorem could not be used if $a,p$ each were one of $14,15$.
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
add a comment |
$begingroup$
Fermat's little theorem says $a^{12}cong1pmod{13}$ for $aneq0pmod{13}$. Thus $2^{12}cong1pmod{13}$. So $x=1$.
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
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$begingroup$
It might worth noting that your solution is easy to check in this case: it is worth memorizing your powers of $2$, or at least certain specific ones (e.g. $2^{10}$). In this case, $2^{12} = 4,096$, which leaves a remainder of $1$ when divided by $13$. In this case, you thus use that to verify you're correct, but it's understandable that this is (a) not the method intended to be used and (b) might not be feasible for the general case.
Anyhow, let us remember Fermat's Little Theorem:
Fermat's Little Theorem: Let $p$ be prime, and $a$ an integer. Then $a^p - a$ is a multiple of $p$. This can be stated symbolically in a few ways: namely:
$$p | (a^p - a) ;;;;;;;;;; frac{a^p - a}{p} = k in mathbb{Z} ;;;;;;;;;; a^p equiv a pmod p$$
Corollary/Implication: If $p not | a$, then this implies $a^{p-1} - 1$ is a multiple of $p$. Again, equivalent formulations:
$$p | (a^{p-1} - 1) ;;;;;;;;;; frac{a^{p-1} - 1}{p} = k in mathbb{Z} ;;;;;;;;;; a^{p-1} equiv 1 pmod p$$
We use this latter form of Fermat's Little Theorem in this problem, with $a = 2, p = 13$.
Then by Fermat's Little Theorem, since $p = 13$ is a prime which does divide $a = 2$, we can claim the following:
$$2^{13-1} equiv 1 pmod {13} iff 2^{12} equiv 1 pmod {13}$$
When $mathbb{Z}_{13}$ denotes the integers mod $13$, then it is clear $1 in mathbb{Z}_{13}$, and thus $x = 1$ in this problem.
Your proof is more or less right, but Fermat's Little Theorem is not used correctly. Having $gcd(2,13) = 1$ is not a sufficient condition for the corollary of Fermat's Little Theorem: one of them must also be prime. (Then $gcd(a,p) = 1$ would imply the necessary condition of non-divisibility.) As an example, $gcd(15, 14) = 1$ but neither are prime, which means Fermat's Little Theorem could not be used if $a,p$ each were one of $14,15$.
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
add a comment |
$begingroup$
It might worth noting that your solution is easy to check in this case: it is worth memorizing your powers of $2$, or at least certain specific ones (e.g. $2^{10}$). In this case, $2^{12} = 4,096$, which leaves a remainder of $1$ when divided by $13$. In this case, you thus use that to verify you're correct, but it's understandable that this is (a) not the method intended to be used and (b) might not be feasible for the general case.
Anyhow, let us remember Fermat's Little Theorem:
Fermat's Little Theorem: Let $p$ be prime, and $a$ an integer. Then $a^p - a$ is a multiple of $p$. This can be stated symbolically in a few ways: namely:
$$p | (a^p - a) ;;;;;;;;;; frac{a^p - a}{p} = k in mathbb{Z} ;;;;;;;;;; a^p equiv a pmod p$$
Corollary/Implication: If $p not | a$, then this implies $a^{p-1} - 1$ is a multiple of $p$. Again, equivalent formulations:
$$p | (a^{p-1} - 1) ;;;;;;;;;; frac{a^{p-1} - 1}{p} = k in mathbb{Z} ;;;;;;;;;; a^{p-1} equiv 1 pmod p$$
We use this latter form of Fermat's Little Theorem in this problem, with $a = 2, p = 13$.
Then by Fermat's Little Theorem, since $p = 13$ is a prime which does divide $a = 2$, we can claim the following:
$$2^{13-1} equiv 1 pmod {13} iff 2^{12} equiv 1 pmod {13}$$
When $mathbb{Z}_{13}$ denotes the integers mod $13$, then it is clear $1 in mathbb{Z}_{13}$, and thus $x = 1$ in this problem.
Your proof is more or less right, but Fermat's Little Theorem is not used correctly. Having $gcd(2,13) = 1$ is not a sufficient condition for the corollary of Fermat's Little Theorem: one of them must also be prime. (Then $gcd(a,p) = 1$ would imply the necessary condition of non-divisibility.) As an example, $gcd(15, 14) = 1$ but neither are prime, which means Fermat's Little Theorem could not be used if $a,p$ each were one of $14,15$.
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
add a comment |
$begingroup$
It might worth noting that your solution is easy to check in this case: it is worth memorizing your powers of $2$, or at least certain specific ones (e.g. $2^{10}$). In this case, $2^{12} = 4,096$, which leaves a remainder of $1$ when divided by $13$. In this case, you thus use that to verify you're correct, but it's understandable that this is (a) not the method intended to be used and (b) might not be feasible for the general case.
Anyhow, let us remember Fermat's Little Theorem:
Fermat's Little Theorem: Let $p$ be prime, and $a$ an integer. Then $a^p - a$ is a multiple of $p$. This can be stated symbolically in a few ways: namely:
$$p | (a^p - a) ;;;;;;;;;; frac{a^p - a}{p} = k in mathbb{Z} ;;;;;;;;;; a^p equiv a pmod p$$
Corollary/Implication: If $p not | a$, then this implies $a^{p-1} - 1$ is a multiple of $p$. Again, equivalent formulations:
$$p | (a^{p-1} - 1) ;;;;;;;;;; frac{a^{p-1} - 1}{p} = k in mathbb{Z} ;;;;;;;;;; a^{p-1} equiv 1 pmod p$$
We use this latter form of Fermat's Little Theorem in this problem, with $a = 2, p = 13$.
Then by Fermat's Little Theorem, since $p = 13$ is a prime which does divide $a = 2$, we can claim the following:
$$2^{13-1} equiv 1 pmod {13} iff 2^{12} equiv 1 pmod {13}$$
When $mathbb{Z}_{13}$ denotes the integers mod $13$, then it is clear $1 in mathbb{Z}_{13}$, and thus $x = 1$ in this problem.
Your proof is more or less right, but Fermat's Little Theorem is not used correctly. Having $gcd(2,13) = 1$ is not a sufficient condition for the corollary of Fermat's Little Theorem: one of them must also be prime. (Then $gcd(a,p) = 1$ would imply the necessary condition of non-divisibility.) As an example, $gcd(15, 14) = 1$ but neither are prime, which means Fermat's Little Theorem could not be used if $a,p$ each were one of $14,15$.
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
It might worth noting that your solution is easy to check in this case: it is worth memorizing your powers of $2$, or at least certain specific ones (e.g. $2^{10}$). In this case, $2^{12} = 4,096$, which leaves a remainder of $1$ when divided by $13$. In this case, you thus use that to verify you're correct, but it's understandable that this is (a) not the method intended to be used and (b) might not be feasible for the general case.
Anyhow, let us remember Fermat's Little Theorem:
Fermat's Little Theorem: Let $p$ be prime, and $a$ an integer. Then $a^p - a$ is a multiple of $p$. This can be stated symbolically in a few ways: namely:
$$p | (a^p - a) ;;;;;;;;;; frac{a^p - a}{p} = k in mathbb{Z} ;;;;;;;;;; a^p equiv a pmod p$$
Corollary/Implication: If $p not | a$, then this implies $a^{p-1} - 1$ is a multiple of $p$. Again, equivalent formulations:
$$p | (a^{p-1} - 1) ;;;;;;;;;; frac{a^{p-1} - 1}{p} = k in mathbb{Z} ;;;;;;;;;; a^{p-1} equiv 1 pmod p$$
We use this latter form of Fermat's Little Theorem in this problem, with $a = 2, p = 13$.
Then by Fermat's Little Theorem, since $p = 13$ is a prime which does divide $a = 2$, we can claim the following:
$$2^{13-1} equiv 1 pmod {13} iff 2^{12} equiv 1 pmod {13}$$
When $mathbb{Z}_{13}$ denotes the integers mod $13$, then it is clear $1 in mathbb{Z}_{13}$, and thus $x = 1$ in this problem.
Your proof is more or less right, but Fermat's Little Theorem is not used correctly. Having $gcd(2,13) = 1$ is not a sufficient condition for the corollary of Fermat's Little Theorem: one of them must also be prime. (Then $gcd(a,p) = 1$ would imply the necessary condition of non-divisibility.) As an example, $gcd(15, 14) = 1$ but neither are prime, which means Fermat's Little Theorem could not be used if $a,p$ each were one of $14,15$.
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
edited Jan 10 at 0:06
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
answered Jan 4 at 1:11
Eevee TrainerEevee Trainer
5,4791936
5,4791936
add a comment |
add a comment |
$begingroup$
Fermat's little theorem says $a^{12}cong1pmod{13}$ for $aneq0pmod{13}$. Thus $2^{12}cong1pmod{13}$. So $x=1$.
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
$endgroup$
add a comment |
$begingroup$
Fermat's little theorem says $a^{12}cong1pmod{13}$ for $aneq0pmod{13}$. Thus $2^{12}cong1pmod{13}$. So $x=1$.
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
$endgroup$
add a comment |
$begingroup$
Fermat's little theorem says $a^{12}cong1pmod{13}$ for $aneq0pmod{13}$. Thus $2^{12}cong1pmod{13}$. So $x=1$.
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
$endgroup$
Fermat's little theorem says $a^{12}cong1pmod{13}$ for $aneq0pmod{13}$. Thus $2^{12}cong1pmod{13}$. So $x=1$.
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
answered Jan 10 at 0:46
Chris CusterChris Custer
11.4k3824
11.4k3824
add a comment |
add a comment |
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You're not using Fermat correctly (well you are, but not for the reason you think you are) : the reason you can use it is because 13 is prime, and because 13 does not divide 2, not only because $gcd(2,13)=1$
$endgroup$
– Astyx
Nov 17 '16 at 17:51
1
$begingroup$
@Astyx If $p$ is prime, $ainmathbb Z^+$, then $pnmid a$ is equivalent to $gcd(p,a)=1$.
$endgroup$
– user236182
Nov 17 '16 at 17:52
2
$begingroup$
@user236182 I absolutely agree, hence the "not only because ..." :)
$endgroup$
– Astyx
Nov 17 '16 at 17:55
$begingroup$
Thank you very much for these info! Is the rest really fine how I wrote it? It seems like a strange theorem to me and the way it's written. Really correct?
$endgroup$
– cnmesr
Nov 17 '16 at 17:57
$begingroup$
It's correct.${}$
$endgroup$
– user236182
Nov 17 '16 at 17:59