Any other quick way to find the remainder? [closed]
$begingroup$
I want to find
$$19^{13}operatorname{mod}(2537).$$
Is there quick way to find the remainder?
elementary-number-theory
edited Dec 15 '18 at 22:22
Namaste
1
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
closed as off-topic by Saad, Davide Giraudo, user10354138, Namaste, Paul Frost Dec 15 '18 at 23:11
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." – Saad, Davide Giraudo, user10354138, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I want to find
$$19^{13}operatorname{mod}(2537).$$
Is there quick way to find the remainder?
elementary-number-theory
edited Dec 15 '18 at 22:22
Namaste
1
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
closed as off-topic by Saad, Davide Giraudo, user10354138, Namaste, Paul Frost Dec 15 '18 at 23:11
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." – Saad, Davide Giraudo, user10354138, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
$$(19,43)=1=(19,59), 2537=43cdot59$$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:56
2
$begingroup$
Expounding on what Peter said, since $19$ is prime, only $1, 19, 19^2, 19^3, ..., 19^{13}$ divide $19^{13}$, and $19$ and any power of it does not divide $2537$.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 9:57
$begingroup$
@ Peter @ Eevee Trainer @ lab bhattacharjee Yes, my question is wrong because my calculation is incorrect. Now I know the answer. Thanks.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:47
add a comment |
$begingroup$
I want to find
$$19^{13}operatorname{mod}(2537).$$
Is there quick way to find the remainder?
elementary-number-theory
edited Dec 15 '18 at 22:22
Namaste
1
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
I want to find
$$19^{13}operatorname{mod}(2537).$$
Is there quick way to find the remainder?
elementary-number-theory
elementary-number-theory
edited Dec 15 '18 at 22:22
Namaste
1
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
edited Dec 15 '18 at 22:22
Namaste
1
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
edited Dec 15 '18 at 22:22
Namaste
1
edited Dec 15 '18 at 22:22
Namaste
1
edited Dec 15 '18 at 22:22
Namaste
1
1
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
asked Dec 15 '18 at 9:54
Ongky Denny WijayaOngky Denny Wijaya
3748
3748
closed as off-topic by Saad, Davide Giraudo, user10354138, Namaste, Paul Frost Dec 15 '18 at 23:11
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." – Saad, Davide Giraudo, user10354138, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Davide Giraudo, user10354138, Namaste, Paul Frost Dec 15 '18 at 23:11
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." – Saad, Davide Giraudo, user10354138, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
$$(19,43)=1=(19,59), 2537=43cdot59$$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:56
2
$begingroup$
Expounding on what Peter said, since $19$ is prime, only $1, 19, 19^2, 19^3, ..., 19^{13}$ divide $19^{13}$, and $19$ and any power of it does not divide $2537$.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 9:57
$begingroup$
@ Peter @ Eevee Trainer @ lab bhattacharjee Yes, my question is wrong because my calculation is incorrect. Now I know the answer. Thanks.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:47
add a comment |
4
$begingroup$
$$(19,43)=1=(19,59), 2537=43cdot59$$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:56
2
$begingroup$
Expounding on what Peter said, since $19$ is prime, only $1, 19, 19^2, 19^3, ..., 19^{13}$ divide $19^{13}$, and $19$ and any power of it does not divide $2537$.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 9:57
$begingroup$
@ Peter @ Eevee Trainer @ lab bhattacharjee Yes, my question is wrong because my calculation is incorrect. Now I know the answer. Thanks.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:47
4
4
$begingroup$
$$(19,43)=1=(19,59), 2537=43cdot59$$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:56
$begingroup$
$$(19,43)=1=(19,59), 2537=43cdot59$$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:56
2
2
$begingroup$
Expounding on what Peter said, since $19$ is prime, only $1, 19, 19^2, 19^3, ..., 19^{13}$ divide $19^{13}$, and $19$ and any power of it does not divide $2537$.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 9:57
$begingroup$
Expounding on what Peter said, since $19$ is prime, only $1, 19, 19^2, 19^3, ..., 19^{13}$ divide $19^{13}$, and $19$ and any power of it does not divide $2537$.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 9:57
$begingroup$
@ Peter @ Eevee Trainer @ lab bhattacharjee Yes, my question is wrong because my calculation is incorrect. Now I know the answer. Thanks.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:47
$begingroup$
@ Peter @ Eevee Trainer @ lab bhattacharjee Yes, my question is wrong because my calculation is incorrect. Now I know the answer. Thanks.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT:
Write $19^{13}=(19^3)^4cdot19$.
Note that $19^3=6859equiv-752pmod{2537}$.
Write $(-752)^4$ as $((-752)^2)^2$ and evaluate the inner bracket to simplify.
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
add a comment |
$begingroup$
begin{eqnarray}
19^3=6859&equiv& 1785(operatorname{mod}2537)\
19^6=left(19^3right)^2equiv (1785)^2=3186225&equiv& 2290(operatorname{mod}2537)\
19^{12}=left(19^6right)^2equiv (2290)^2 = 5244100&equiv& 121(operatorname{mod}2537)\
19^{13}=19^{12}cdot19&equiv& 121cdot 19=2299(operatorname{mod}2537)
end{eqnarray}
So the remainder is 2299. Is this correct answer?
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
1
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
Write $19^{13}=(19^3)^4cdot19$.
Note that $19^3=6859equiv-752pmod{2537}$.
Write $(-752)^4$ as $((-752)^2)^2$ and evaluate the inner bracket to simplify.
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
add a comment |
$begingroup$
HINT:
Write $19^{13}=(19^3)^4cdot19$.
Note that $19^3=6859equiv-752pmod{2537}$.
Write $(-752)^4$ as $((-752)^2)^2$ and evaluate the inner bracket to simplify.
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
add a comment |
$begingroup$
HINT:
Write $19^{13}=(19^3)^4cdot19$.
Note that $19^3=6859equiv-752pmod{2537}$.
Write $(-752)^4$ as $((-752)^2)^2$ and evaluate the inner bracket to simplify.
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
HINT:
Write $19^{13}=(19^3)^4cdot19$.
Note that $19^3=6859equiv-752pmod{2537}$.
Write $(-752)^4$ as $((-752)^2)^2$ and evaluate the inner bracket to simplify.
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
answered Dec 15 '18 at 10:06
TheSimpliFireTheSimpliFire
12.7k62461
12.7k62461
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
add a comment |
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Please check my answer, is it correct answer?
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:43
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
Quite rather a PSQ, don't you think?
$endgroup$
– Namaste
Dec 15 '18 at 22:23
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
$begingroup$
@amWhy They originally had some working but it was pointed out that it was wrong so they deleted it. However, they have provided their own correct answer below mine so I won't be too harsh about it.
$endgroup$
– TheSimpliFire
Dec 16 '18 at 8:56
add a comment |
$begingroup$
begin{eqnarray}
19^3=6859&equiv& 1785(operatorname{mod}2537)\
19^6=left(19^3right)^2equiv (1785)^2=3186225&equiv& 2290(operatorname{mod}2537)\
19^{12}=left(19^6right)^2equiv (2290)^2 = 5244100&equiv& 121(operatorname{mod}2537)\
19^{13}=19^{12}cdot19&equiv& 121cdot 19=2299(operatorname{mod}2537)
end{eqnarray}
So the remainder is 2299. Is this correct answer?
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
1
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
add a comment |
$begingroup$
begin{eqnarray}
19^3=6859&equiv& 1785(operatorname{mod}2537)\
19^6=left(19^3right)^2equiv (1785)^2=3186225&equiv& 2290(operatorname{mod}2537)\
19^{12}=left(19^6right)^2equiv (2290)^2 = 5244100&equiv& 121(operatorname{mod}2537)\
19^{13}=19^{12}cdot19&equiv& 121cdot 19=2299(operatorname{mod}2537)
end{eqnarray}
So the remainder is 2299. Is this correct answer?
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
1
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
add a comment |
$begingroup$
begin{eqnarray}
19^3=6859&equiv& 1785(operatorname{mod}2537)\
19^6=left(19^3right)^2equiv (1785)^2=3186225&equiv& 2290(operatorname{mod}2537)\
19^{12}=left(19^6right)^2equiv (2290)^2 = 5244100&equiv& 121(operatorname{mod}2537)\
19^{13}=19^{12}cdot19&equiv& 121cdot 19=2299(operatorname{mod}2537)
end{eqnarray}
So the remainder is 2299. Is this correct answer?
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
$endgroup$
begin{eqnarray}
19^3=6859&equiv& 1785(operatorname{mod}2537)\
19^6=left(19^3right)^2equiv (1785)^2=3186225&equiv& 2290(operatorname{mod}2537)\
19^{12}=left(19^6right)^2equiv (2290)^2 = 5244100&equiv& 121(operatorname{mod}2537)\
19^{13}=19^{12}cdot19&equiv& 121cdot 19=2299(operatorname{mod}2537)
end{eqnarray}
So the remainder is 2299. Is this correct answer?
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
answered Dec 15 '18 at 10:41
Ongky Denny WijayaOngky Denny Wijaya
3748
3748
1
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
add a comment |
1
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
1
1
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Yes${}{}{}{}{}{}$
$endgroup$
– TheSimpliFire
Dec 15 '18 at 10:55
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
$begingroup$
Thank you very much for your hint, @ TheSimpliFire.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 11:00
add a comment |
4
$begingroup$
$$(19,43)=1=(19,59), 2537=43cdot59$$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:56
2
$begingroup$
Expounding on what Peter said, since $19$ is prime, only $1, 19, 19^2, 19^3, ..., 19^{13}$ divide $19^{13}$, and $19$ and any power of it does not divide $2537$.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 9:57
$begingroup$
@ Peter @ Eevee Trainer @ lab bhattacharjee Yes, my question is wrong because my calculation is incorrect. Now I know the answer. Thanks.
$endgroup$
– Ongky Denny Wijaya
Dec 15 '18 at 10:47