What will be the remainder if $7^{101}$ mod $5$? [closed]
$begingroup$
$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$
What is the next step to find solution?
Regards
modular-arithmetic
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
asked Dec 14 '18 at 16:16
MarcinMarcin
1
$endgroup$
closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$
What is the next step to find solution?
Regards
modular-arithmetic
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
asked Dec 14 '18 at 16:16
MarcinMarcin
1
$endgroup$
closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18
2
$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27
$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27
1
$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55
$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13
add a comment |
$begingroup$
$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$
What is the next step to find solution?
Regards
modular-arithmetic
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
asked Dec 14 '18 at 16:16
MarcinMarcin
1
$endgroup$
$$7^{101} mod 5 = 49^{100} mod 5 = (49^2)^{50} mod 5$$
What is the next step to find solution?
Regards
modular-arithmetic
modular-arithmetic
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
asked Dec 14 '18 at 16:16
MarcinMarcin
1
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
asked Dec 14 '18 at 16:16
MarcinMarcin
1
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
edited Dec 14 '18 at 18:54
Alex Kruckman
27.4k32657
27.4k32657
asked Dec 14 '18 at 16:16
MarcinMarcin
1
asked Dec 14 '18 at 16:16
MarcinMarcin
1
asked Dec 14 '18 at 16:16
MarcinMarcin
1
1
closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED Dec 14 '18 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Arnaud D., amWhy, user10354138, Math_QED
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18
2
$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27
$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27
1
$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55
$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13
add a comment |
1
$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18
2
$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27
$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27
1
$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55
$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13
1
1
$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18
$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18
2
2
$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27
$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27
$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27
$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27
1
1
$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55
$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55
$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13
$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is something wrong in your first step, indeed
$$7^{101} not equiv 49^{100} mod 5 ldots $$
More simply use that
$$7^2 equiv 49 equiv -1 pmod 5$$
therefore
$$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.
$7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.
answered Dec 14 '18 at 18:25
yavaryavar
993
$endgroup$
$begingroup$
usepmod{5}to render modulo.
$endgroup$
– zwim
Jan 7 at 21:23
add a comment |
$begingroup$
What if you do not see a cute way to do this?
There is a more systematic way to do it.
First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is something wrong in your first step, indeed
$$7^{101} not equiv 49^{100} mod 5 ldots $$
More simply use that
$$7^2 equiv 49 equiv -1 pmod 5$$
therefore
$$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
There is something wrong in your first step, indeed
$$7^{101} not equiv 49^{100} mod 5 ldots $$
More simply use that
$$7^2 equiv 49 equiv -1 pmod 5$$
therefore
$$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
There is something wrong in your first step, indeed
$$7^{101} not equiv 49^{100} mod 5 ldots $$
More simply use that
$$7^2 equiv 49 equiv -1 pmod 5$$
therefore
$$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
$endgroup$
There is something wrong in your first step, indeed
$$7^{101} not equiv 49^{100} mod 5 ldots $$
More simply use that
$$7^2 equiv 49 equiv -1 pmod 5$$
therefore
$$7^{101} equiv 7cdot7^{100} equiv 7cdot (7^{2})^{50} equiv ,? pmod 5$$
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
edited Dec 14 '18 at 16:25
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
answered Dec 14 '18 at 16:20
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.
$7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.
answered Dec 14 '18 at 18:25
yavaryavar
993
$endgroup$
$begingroup$
usepmod{5}to render modulo.
$endgroup$
– zwim
Jan 7 at 21:23
add a comment |
$begingroup$
By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.
$7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.
answered Dec 14 '18 at 18:25
yavaryavar
993
$endgroup$
$begingroup$
usepmod{5}to render modulo.
$endgroup$
– zwim
Jan 7 at 21:23
add a comment |
$begingroup$
By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.
$7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.
answered Dec 14 '18 at 18:25
yavaryavar
993
$endgroup$
By Carmichael function if $(a,m)=1$, then $a^{k lambda (M)}equiv 1 (mod m)$. That $lambda (5)=4$. So $7^{4k}equiv 1 (mod 5)$.
$7^{100}equiv 1 (mod 5)$ then $7^{101}equiv 2 (mod 5)$.
answered Dec 14 '18 at 18:25
yavaryavar
993
edited Dec 15 '18 at 18:14
answered Dec 14 '18 at 18:25
yavaryavar
993
answered Dec 14 '18 at 18:25
yavaryavar
993
answered Dec 14 '18 at 18:25
yavaryavar
993
993
$begingroup$
usepmod{5}to render modulo.
$endgroup$
– zwim
Jan 7 at 21:23
add a comment |
$begingroup$
usepmod{5}to render modulo.
$endgroup$
– zwim
Jan 7 at 21:23
$begingroup$
use
pmod{5} to render modulo.$endgroup$
– zwim
Jan 7 at 21:23
$begingroup$
use
pmod{5} to render modulo.$endgroup$
– zwim
Jan 7 at 21:23
add a comment |
$begingroup$
What if you do not see a cute way to do this?
There is a more systematic way to do it.
First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
$endgroup$
add a comment |
$begingroup$
What if you do not see a cute way to do this?
There is a more systematic way to do it.
First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
$endgroup$
add a comment |
$begingroup$
What if you do not see a cute way to do this?
There is a more systematic way to do it.
First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
$endgroup$
What if you do not see a cute way to do this?
There is a more systematic way to do it.
First, $7^xpmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}pmod 5$.
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
answered Dec 14 '18 at 19:00
GEdgarGEdgar
62.5k267171
62.5k267171
add a comment |
add a comment |
1
$begingroup$
$large bmod 5!: 7equiv 2, color{#c00}{2^4equiv 1},Rightarrow, 2^{4n}equiv (color{#c00}{2^4})^nequiv color{#c00}1^nequiv 1 $
$endgroup$
– Bill Dubuque
Dec 14 '18 at 16:18
2
$begingroup$
How did you get from $7^{101}$ to $49^{100}$?
$endgroup$
– farruhota
Dec 14 '18 at 16:27
$begingroup$
You are wrong in saying $7^{101}equiv 49^{100}$ you need $7cdot 49^{50}$ - just take care when manipulating the powers.
$endgroup$
– Mark Bennet
Dec 14 '18 at 16:27
1
$begingroup$
Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)!
$endgroup$
– Alex Kruckman
Dec 14 '18 at 18:55
$begingroup$
This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 6:13