Find a $x$ such that $2^{2015}xequiv 1 pmod{13}$ [on hold]

up vote
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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

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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

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

share|cite|improve this question

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

share|cite|improve this question

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

share|cite|improve this question

edited yesterday

Bill Dubuque

206k29189621

asked yesterday

Marko Škorić

58110

share|cite|improve this question

edited yesterday

Bill Dubuque

206k29189621

asked yesterday

Marko Škorić

58110

share|cite|improve this question

share|cite|improve this question

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

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.

share|cite|improve this answer

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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.

share|cite|improve this answer

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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.

share|cite|improve this answer

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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.

share|cite|improve this answer

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.

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.

share|cite|improve this answer

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.

share|cite|improve this answer

share|cite|improve this answer

edited yesterday

edited yesterday

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 | 

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