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Find a $x$ such that $2^{2015}xequiv 1 pmod{13}$ [on hold]
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Since 13 is prime number using little Fermat's theorem $2^{12}equiv 1 pmod {13}$ then $2^{2015}equiv 2^{12cdot167+11}equiv 2^{11} pmod{13}$ then $2^{2015} x equiv 2^{11} x equiv 1 pmod{13}$ so then $xequiv 2$, is this ok?
discrete-mathematics modular-arithmetic divisibility
edited yesterday
Bill Dubuque
206k29189621
asked yesterday
Marko Škorić
58110
put on hold as unclear what you're asking by Rushabh Mehta, Xander Henderson, Servaes, amWhy, José Carlos Santos yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-2
down vote
favorite
Since 13 is prime number using little Fermat's theorem $2^{12}equiv 1 pmod {13}$ then $2^{2015}equiv 2^{12cdot167+11}equiv 2^{11} pmod{13}$ then $2^{2015} x equiv 2^{11} x equiv 1 pmod{13}$ so then $xequiv 2$, is this ok?
discrete-mathematics modular-arithmetic divisibility
edited yesterday
Bill Dubuque
206k29189621
asked yesterday
Marko Škorić
58110
put on hold as unclear what you're asking by Rushabh Mehta, Xander Henderson, Servaes, amWhy, José Carlos Santos yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
There is no $x$ in $2^{2015}equiv 1pmod{13}$. Given that $2^{2015}equiv7pmod{13}$, there is no such $x$ at all.
– Servaes
yesterday
2
So you mean "Find"? or are we expected to "Wind" $x$ about something? or is it just windy where you are?
– Xander Henderson
yesterday
I think the OP meant to say find $x$ such that $2^{2015} x equiv 1 pmod{13}$.
– Math Lover
yesterday
sorry I did not saw that I write wind
– Marko Škorić
yesterday
To check that $x=2$ works is the same as checking $2^{2016}equiv 1 pmod {13}$, but that follows from the fact that $12,|,2016$.
– lulu
yesterday
|
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Since 13 is prime number using little Fermat's theorem $2^{12}equiv 1 pmod {13}$ then $2^{2015}equiv 2^{12cdot167+11}equiv 2^{11} pmod{13}$ then $2^{2015} x equiv 2^{11} x equiv 1 pmod{13}$ so then $xequiv 2$, is this ok?
discrete-mathematics modular-arithmetic divisibility
edited yesterday
Bill Dubuque
206k29189621
asked yesterday
Marko Škorić
58110
Since 13 is prime number using little Fermat's theorem $2^{12}equiv 1 pmod {13}$ then $2^{2015}equiv 2^{12cdot167+11}equiv 2^{11} pmod{13}$ then $2^{2015} x equiv 2^{11} x equiv 1 pmod{13}$ so then $xequiv 2$, is this ok?
discrete-mathematics modular-arithmetic divisibility
discrete-mathematics modular-arithmetic divisibility
edited yesterday
Bill Dubuque
206k29189621
asked yesterday
Marko Škorić
58110
edited yesterday
Bill Dubuque
206k29189621
asked yesterday
Marko Škorić
58110
edited yesterday
Bill Dubuque
206k29189621
edited yesterday
Bill Dubuque
206k29189621
edited yesterday
Bill Dubuque
206k29189621
206k29189621
asked yesterday
Marko Škorić
58110
asked yesterday
Marko Škorić
58110
asked yesterday
Marko Škorić
58110
58110
put on hold as unclear what you're asking by Rushabh Mehta, Xander Henderson, Servaes, amWhy, José Carlos Santos yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Rushabh Mehta, Xander Henderson, Servaes, amWhy, José Carlos Santos yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
There is no $x$ in $2^{2015}equiv 1pmod{13}$. Given that $2^{2015}equiv7pmod{13}$, there is no such $x$ at all.
– Servaes
yesterday
2
So you mean "Find"? or are we expected to "Wind" $x$ about something? or is it just windy where you are?
– Xander Henderson
yesterday
I think the OP meant to say find $x$ such that $2^{2015} x equiv 1 pmod{13}$.
– Math Lover
yesterday
sorry I did not saw that I write wind
– Marko Škorić
yesterday
To check that $x=2$ works is the same as checking $2^{2016}equiv 1 pmod {13}$, but that follows from the fact that $12,|,2016$.
– lulu
yesterday
|
show 2 more comments
3
There is no $x$ in $2^{2015}equiv 1pmod{13}$. Given that $2^{2015}equiv7pmod{13}$, there is no such $x$ at all.
– Servaes
yesterday
2
So you mean "Find"? or are we expected to "Wind" $x$ about something? or is it just windy where you are?
– Xander Henderson
yesterday
I think the OP meant to say find $x$ such that $2^{2015} x equiv 1 pmod{13}$.
– Math Lover
yesterday
sorry I did not saw that I write wind
– Marko Škorić
yesterday
To check that $x=2$ works is the same as checking $2^{2016}equiv 1 pmod {13}$, but that follows from the fact that $12,|,2016$.
– lulu
yesterday
3
3
There is no $x$ in $2^{2015}equiv 1pmod{13}$. Given that $2^{2015}equiv7pmod{13}$, there is no such $x$ at all.
– Servaes
yesterday
There is no $x$ in $2^{2015}equiv 1pmod{13}$. Given that $2^{2015}equiv7pmod{13}$, there is no such $x$ at all.
– Servaes
yesterday
2
2
So you mean "Find"? or are we expected to "Wind" $x$ about something? or is it just windy where you are?
– Xander Henderson
yesterday
So you mean "Find"? or are we expected to "Wind" $x$ about something? or is it just windy where you are?
– Xander Henderson
yesterday
I think the OP meant to say find $x$ such that $2^{2015} x equiv 1 pmod{13}$.
– Math Lover
yesterday
I think the OP meant to say find $x$ such that $2^{2015} x equiv 1 pmod{13}$.
– Math Lover
yesterday
sorry I did not saw that I write wind
– Marko Škorić
yesterday
sorry I did not saw that I write wind
– Marko Škorić
yesterday
To check that $x=2$ works is the same as checking $2^{2016}equiv 1 pmod {13}$, but that follows from the fact that $12,|,2016$.
– lulu
yesterday
To check that $x=2$ works is the same as checking $2^{2016}equiv 1 pmod {13}$, but that follows from the fact that $12,|,2016$.
– lulu
yesterday
|
show 2 more comments
1 Answer
1
active
oldest
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up vote
2
down vote
accepted
By Fermat's theorem,
$$2^{12}equiv 1pmod{13}$$
Observe that, ${12}mid {2016}$
Therefore, $$ 2^{2016}equiv 1pmod{13}$$
Which is same as $$2^{2015}times 2equiv 1pmod{13}$$
So $x= 2$ is a solution of the congruence.
answered yesterday
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
1
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By Fermat's theorem,
$$2^{12}equiv 1pmod{13}$$
Observe that, ${12}mid {2016}$
Therefore, $$ 2^{2016}equiv 1pmod{13}$$
Which is same as $$2^{2015}times 2equiv 1pmod{13}$$
So $x= 2$ is a solution of the congruence.
answered yesterday
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
1
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
add a comment |
up vote
2
down vote
accepted
By Fermat's theorem,
$$2^{12}equiv 1pmod{13}$$
Observe that, ${12}mid {2016}$
Therefore, $$ 2^{2016}equiv 1pmod{13}$$
Which is same as $$2^{2015}times 2equiv 1pmod{13}$$
So $x= 2$ is a solution of the congruence.
answered yesterday
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
1
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By Fermat's theorem,
$$2^{12}equiv 1pmod{13}$$
Observe that, ${12}mid {2016}$
Therefore, $$ 2^{2016}equiv 1pmod{13}$$
Which is same as $$2^{2015}times 2equiv 1pmod{13}$$
So $x= 2$ is a solution of the congruence.
answered yesterday
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
By Fermat's theorem,
$$2^{12}equiv 1pmod{13}$$
Observe that, ${12}mid {2016}$
Therefore, $$ 2^{2016}equiv 1pmod{13}$$
Which is same as $$2^{2015}times 2equiv 1pmod{13}$$
So $x= 2$ is a solution of the congruence.
answered yesterday
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Samurai
987
answered yesterday
Samurai
987
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
1
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
add a comment |
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
1
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
Also note that $2^{2015}equiv 7pmod{13}$ . So the above congruence is equivalent to $7xequiv 1pmod{13}$ ,which is a linear congruence and easier to solve.
– Samurai
yesterday
1
1
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
You should write "so $xequiv 2$ is a solution of the congruence". Otherwise it is not clear which direction "so" denotes (clarity is essential here since often necessary and sufficient conditions are mixed up).
– Bill Dubuque
yesterday
add a comment |
3
There is no $x$ in $2^{2015}equiv 1pmod{13}$. Given that $2^{2015}equiv7pmod{13}$, there is no such $x$ at all.
– Servaes
yesterday
2
So you mean "Find"? or are we expected to "Wind" $x$ about something? or is it just windy where you are?
– Xander Henderson
yesterday
I think the OP meant to say find $x$ such that $2^{2015} x equiv 1 pmod{13}$.
– Math Lover
yesterday
sorry I did not saw that I write wind
– Marko Škorić
yesterday
To check that $x=2$ works is the same as checking $2^{2016}equiv 1 pmod {13}$, but that follows from the fact that $12,|,2016$.
– lulu
yesterday