Determine the remainder we get if we divide $799^{801}$ by $264$ [closed]
1
$begingroup$
Determine the remainder we get if we divide $799^{801}$ by $264$.
I need clear steps.
elementary-number-theory modular-arithmetic
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
$endgroup$
closed as off-topic by amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen Dec 13 '18 at 17:29
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." – amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
1
$begingroup$
Determine the remainder we get if we divide $799^{801}$ by $264$.
I need clear steps.
elementary-number-theory modular-arithmetic
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
$endgroup$
closed as off-topic by amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen Dec 13 '18 at 17:29
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." – amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What have you tried? Do you know Fermat's Little Theorem?
$endgroup$
– lulu
Dec 13 '18 at 16:44
$begingroup$
I can get 799^800*&799≡1(264). Someone told me the answer is 799≡7(264). I don't understand where is from the 7.
$endgroup$
– Y.xin
Dec 13 '18 at 16:47
$begingroup$
$799 = 264cdot 3 + 7$ therefore $799^{801} =(264cdot 3 + 7)^{801}$. The only term in that huge expansion that isnt already divisible by 264 is $7^{801}$
$endgroup$
– David Diaz
Dec 13 '18 at 16:53
$begingroup$
You do need to give a brief description of what class you are in and what you have studied so for. If you know the Eulers Theorem it's straightforward.
$endgroup$
– fleablood
Dec 13 '18 at 17:11
add a comment |
1
1
1
$begingroup$
Determine the remainder we get if we divide $799^{801}$ by $264$.
I need clear steps.
elementary-number-theory modular-arithmetic
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
$endgroup$
Determine the remainder we get if we divide $799^{801}$ by $264$.
I need clear steps.
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
edited Dec 13 '18 at 16:48
lab bhattacharjee
226k15157275
226k15157275
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
asked Dec 13 '18 at 16:42
Y.xinY.xin
114
114
closed as off-topic by amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen Dec 13 '18 at 17:29
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." – amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen Dec 13 '18 at 17:29
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." – amWhy, jameselmore, MisterRiemann, Xander Henderson, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What have you tried? Do you know Fermat's Little Theorem?
$endgroup$
– lulu
Dec 13 '18 at 16:44
$begingroup$
I can get 799^800*&799≡1(264). Someone told me the answer is 799≡7(264). I don't understand where is from the 7.
$endgroup$
– Y.xin
Dec 13 '18 at 16:47
$begingroup$
$799 = 264cdot 3 + 7$ therefore $799^{801} =(264cdot 3 + 7)^{801}$. The only term in that huge expansion that isnt already divisible by 264 is $7^{801}$
$endgroup$
– David Diaz
Dec 13 '18 at 16:53
$begingroup$
You do need to give a brief description of what class you are in and what you have studied so for. If you know the Eulers Theorem it's straightforward.
$endgroup$
– fleablood
Dec 13 '18 at 17:11
add a comment |
1
$begingroup$
What have you tried? Do you know Fermat's Little Theorem?
$endgroup$
– lulu
Dec 13 '18 at 16:44
$begingroup$
I can get 799^800*&799≡1(264). Someone told me the answer is 799≡7(264). I don't understand where is from the 7.
$endgroup$
– Y.xin
Dec 13 '18 at 16:47
$begingroup$
$799 = 264cdot 3 + 7$ therefore $799^{801} =(264cdot 3 + 7)^{801}$. The only term in that huge expansion that isnt already divisible by 264 is $7^{801}$
$endgroup$
– David Diaz
Dec 13 '18 at 16:53
$begingroup$
You do need to give a brief description of what class you are in and what you have studied so for. If you know the Eulers Theorem it's straightforward.
$endgroup$
– fleablood
Dec 13 '18 at 17:11
1
1
$begingroup$
What have you tried? Do you know Fermat's Little Theorem?
$endgroup$
– lulu
Dec 13 '18 at 16:44
$begingroup$
What have you tried? Do you know Fermat's Little Theorem?
$endgroup$
– lulu
Dec 13 '18 at 16:44
$begingroup$
I can get 799^800*&799≡1(264). Someone told me the answer is 799≡7(264). I don't understand where is from the 7.
$endgroup$
– Y.xin
Dec 13 '18 at 16:47
$begingroup$
I can get 799^800*&799≡1(264). Someone told me the answer is 799≡7(264). I don't understand where is from the 7.
$endgroup$
– Y.xin
Dec 13 '18 at 16:47
$begingroup$
$799 = 264cdot 3 + 7$ therefore $799^{801} =(264cdot 3 + 7)^{801}$. The only term in that huge expansion that isnt already divisible by 264 is $7^{801}$
$endgroup$
– David Diaz
Dec 13 '18 at 16:53
$begingroup$
$799 = 264cdot 3 + 7$ therefore $799^{801} =(264cdot 3 + 7)^{801}$. The only term in that huge expansion that isnt already divisible by 264 is $7^{801}$
$endgroup$
– David Diaz
Dec 13 '18 at 16:53
$begingroup$
You do need to give a brief description of what class you are in and what you have studied so for. If you know the Eulers Theorem it's straightforward.
$endgroup$
– fleablood
Dec 13 '18 at 17:11
$begingroup$
You do need to give a brief description of what class you are in and what you have studied so for. If you know the Eulers Theorem it's straightforward.
$endgroup$
– fleablood
Dec 13 '18 at 17:11
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
$799equiv7pmod{264},799^nequiv7^n$ for integer $nge0$
Using Carmichael Function, $lambda(264)=$lcm$(lambda(3),lambda(8),lambda(11))=10$
As $(7,264)=1$
$$implies7^{801}equiv7^{801pmod{10}}equiv7^1pmod{264}$$
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
1
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
$799equiv7pmod{264},799^nequiv7^n$ for integer $nge0$
Using Carmichael Function, $lambda(264)=$lcm$(lambda(3),lambda(8),lambda(11))=10$
As $(7,264)=1$
$$implies7^{801}equiv7^{801pmod{10}}equiv7^1pmod{264}$$
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
1
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
add a comment |
2
$begingroup$
$799equiv7pmod{264},799^nequiv7^n$ for integer $nge0$
Using Carmichael Function, $lambda(264)=$lcm$(lambda(3),lambda(8),lambda(11))=10$
As $(7,264)=1$
$$implies7^{801}equiv7^{801pmod{10}}equiv7^1pmod{264}$$
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
1
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
add a comment |
2
2
2
$begingroup$
$799equiv7pmod{264},799^nequiv7^n$ for integer $nge0$
Using Carmichael Function, $lambda(264)=$lcm$(lambda(3),lambda(8),lambda(11))=10$
As $(7,264)=1$
$$implies7^{801}equiv7^{801pmod{10}}equiv7^1pmod{264}$$
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$799equiv7pmod{264},799^nequiv7^n$ for integer $nge0$
Using Carmichael Function, $lambda(264)=$lcm$(lambda(3),lambda(8),lambda(11))=10$
As $(7,264)=1$
$$implies7^{801}equiv7^{801pmod{10}}equiv7^1pmod{264}$$
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 13 '18 at 16:45
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
1
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
add a comment |
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
1
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
Hello, where is from the 7?
$endgroup$
– Y.xin
Dec 13 '18 at 16:48
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, $$799%264=?$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:49
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
$begingroup$
@Y.xin, Are you aware of mathworld.wolfram.com/Congruence.html
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 16:52
1
1
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Getting to $7$ is easy: You take $799$ and subtract $264$ you will have $535$ then subtract $264$ again you will have $271$ one again will get you $7$. In other words 799 =3 x 264 + 7.
$endgroup$
– Yanko
Dec 13 '18 at 16:53
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
$begingroup$
Thanks, I will see.
$endgroup$
– Y.xin
Dec 13 '18 at 16:54
add a comment |
1
$begingroup$
What have you tried? Do you know Fermat's Little Theorem?
$endgroup$
– lulu
Dec 13 '18 at 16:44
$begingroup$
I can get 799^800*&799≡1(264). Someone told me the answer is 799≡7(264). I don't understand where is from the 7.
$endgroup$
– Y.xin
Dec 13 '18 at 16:47
$begingroup$
$799 = 264cdot 3 + 7$ therefore $799^{801} =(264cdot 3 + 7)^{801}$. The only term in that huge expansion that isnt already divisible by 264 is $7^{801}$
$endgroup$
– David Diaz
Dec 13 '18 at 16:53
$begingroup$
You do need to give a brief description of what class you are in and what you have studied so for. If you know the Eulers Theorem it's straightforward.
$endgroup$
– fleablood
Dec 13 '18 at 17:11