How to find least residue with big power? [closed]
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Can someone help me with this equation?
$3^{401}+7^{401}equiv xpmod{500}$
Thank you!
modular-arithmetic
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
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closed as off-topic by Carl Mummert, amWhy, Gibbs, José Carlos Santos, T. Bongers Dec 7 '18 at 1:23
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$begingroup$
Can someone help me with this equation?
$3^{401}+7^{401}equiv xpmod{500}$
Thank you!
modular-arithmetic
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
$endgroup$
closed as off-topic by Carl Mummert, amWhy, Gibbs, José Carlos Santos, T. Bongers Dec 7 '18 at 1:23
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, amWhy, Gibbs, José Carlos Santos, T. Bongers
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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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.
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– Carl Mummert
Dec 6 '18 at 17:50
add a comment |
$begingroup$
Can someone help me with this equation?
$3^{401}+7^{401}equiv xpmod{500}$
Thank you!
modular-arithmetic
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
$endgroup$
Can someone help me with this equation?
$3^{401}+7^{401}equiv xpmod{500}$
Thank you!
modular-arithmetic
modular-arithmetic
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
asked Dec 1 '18 at 23:33
ytsejamytsejam
11
11
closed as off-topic by Carl Mummert, amWhy, Gibbs, José Carlos Santos, T. Bongers Dec 7 '18 at 1:23
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, amWhy, Gibbs, José Carlos Santos, T. Bongers
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, amWhy, Gibbs, José Carlos Santos, T. Bongers Dec 7 '18 at 1:23
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, amWhy, Gibbs, José Carlos Santos, T. Bongers
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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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.
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– Carl Mummert
Dec 6 '18 at 17:50
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1
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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.
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– Carl Mummert
Dec 6 '18 at 17:50
1
1
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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.
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– Carl Mummert
Dec 6 '18 at 17:50
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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.
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– Carl Mummert
Dec 6 '18 at 17:50
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3 Answers
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Hint: $varphi (500)=200$, where $varphi $ is Euler's totient function. Thus, by Euler's theorem, $3^{200}cong7^{200}cong1pmod{500}$.
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
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With the intent of using Euler's theorem ($a^{phi(n)} =1 text{ mod n}$), note that $phi(500)=200$. Then one has
$$3^{401} = 3^{200}*3^{200}*3 implies 3^{401} = 3 text{ mod } 500$$
A similar argument shows that $7^{401} = 7text{ mod 500}$.
So then $3^{401}+7^{401}text{ mod 500} = (3^{401}text{ mod 500})+(7^{401}text{ mod 500}) text{ mod 500} = 10 text{ mod 500} $
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
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Use Euler’s Theorem: If $gcd(a,m) = 1$ then $a^{phi(m)} equiv pmod{m}$, where $phi$ is Euler's totient function. Hence, if $gcd(a,m) = 1$ then $x equiv y pmod{phi(m)} Rightarrow a^x equiv a^y pmod{m}$. We have $phi(500) = 200$, and $401 equiv 1 pmod{200}$, hence $3^{401} + 7^{401} equiv 3 + 7 = 10 pmod{500}$.
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $varphi (500)=200$, where $varphi $ is Euler's totient function. Thus, by Euler's theorem, $3^{200}cong7^{200}cong1pmod{500}$.
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
$endgroup$
add a comment |
$begingroup$
Hint: $varphi (500)=200$, where $varphi $ is Euler's totient function. Thus, by Euler's theorem, $3^{200}cong7^{200}cong1pmod{500}$.
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
$endgroup$
add a comment |
$begingroup$
Hint: $varphi (500)=200$, where $varphi $ is Euler's totient function. Thus, by Euler's theorem, $3^{200}cong7^{200}cong1pmod{500}$.
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
$endgroup$
Hint: $varphi (500)=200$, where $varphi $ is Euler's totient function. Thus, by Euler's theorem, $3^{200}cong7^{200}cong1pmod{500}$.
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
edited Dec 1 '18 at 23:58
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
answered Dec 1 '18 at 23:46
Chris CusterChris Custer
11.2k3824
11.2k3824
add a comment |
add a comment |
$begingroup$
With the intent of using Euler's theorem ($a^{phi(n)} =1 text{ mod n}$), note that $phi(500)=200$. Then one has
$$3^{401} = 3^{200}*3^{200}*3 implies 3^{401} = 3 text{ mod } 500$$
A similar argument shows that $7^{401} = 7text{ mod 500}$.
So then $3^{401}+7^{401}text{ mod 500} = (3^{401}text{ mod 500})+(7^{401}text{ mod 500}) text{ mod 500} = 10 text{ mod 500} $
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
$endgroup$
add a comment |
$begingroup$
With the intent of using Euler's theorem ($a^{phi(n)} =1 text{ mod n}$), note that $phi(500)=200$. Then one has
$$3^{401} = 3^{200}*3^{200}*3 implies 3^{401} = 3 text{ mod } 500$$
A similar argument shows that $7^{401} = 7text{ mod 500}$.
So then $3^{401}+7^{401}text{ mod 500} = (3^{401}text{ mod 500})+(7^{401}text{ mod 500}) text{ mod 500} = 10 text{ mod 500} $
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
$endgroup$
add a comment |
$begingroup$
With the intent of using Euler's theorem ($a^{phi(n)} =1 text{ mod n}$), note that $phi(500)=200$. Then one has
$$3^{401} = 3^{200}*3^{200}*3 implies 3^{401} = 3 text{ mod } 500$$
A similar argument shows that $7^{401} = 7text{ mod 500}$.
So then $3^{401}+7^{401}text{ mod 500} = (3^{401}text{ mod 500})+(7^{401}text{ mod 500}) text{ mod 500} = 10 text{ mod 500} $
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
$endgroup$
With the intent of using Euler's theorem ($a^{phi(n)} =1 text{ mod n}$), note that $phi(500)=200$. Then one has
$$3^{401} = 3^{200}*3^{200}*3 implies 3^{401} = 3 text{ mod } 500$$
A similar argument shows that $7^{401} = 7text{ mod 500}$.
So then $3^{401}+7^{401}text{ mod 500} = (3^{401}text{ mod 500})+(7^{401}text{ mod 500}) text{ mod 500} = 10 text{ mod 500} $
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
answered Dec 1 '18 at 23:48
Theo C.Theo C.
25928
25928
add a comment |
add a comment |
$begingroup$
Use Euler’s Theorem: If $gcd(a,m) = 1$ then $a^{phi(m)} equiv pmod{m}$, where $phi$ is Euler's totient function. Hence, if $gcd(a,m) = 1$ then $x equiv y pmod{phi(m)} Rightarrow a^x equiv a^y pmod{m}$. We have $phi(500) = 200$, and $401 equiv 1 pmod{200}$, hence $3^{401} + 7^{401} equiv 3 + 7 = 10 pmod{500}$.
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
$endgroup$
add a comment |
$begingroup$
Use Euler’s Theorem: If $gcd(a,m) = 1$ then $a^{phi(m)} equiv pmod{m}$, where $phi$ is Euler's totient function. Hence, if $gcd(a,m) = 1$ then $x equiv y pmod{phi(m)} Rightarrow a^x equiv a^y pmod{m}$. We have $phi(500) = 200$, and $401 equiv 1 pmod{200}$, hence $3^{401} + 7^{401} equiv 3 + 7 = 10 pmod{500}$.
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
$endgroup$
add a comment |
$begingroup$
Use Euler’s Theorem: If $gcd(a,m) = 1$ then $a^{phi(m)} equiv pmod{m}$, where $phi$ is Euler's totient function. Hence, if $gcd(a,m) = 1$ then $x equiv y pmod{phi(m)} Rightarrow a^x equiv a^y pmod{m}$. We have $phi(500) = 200$, and $401 equiv 1 pmod{200}$, hence $3^{401} + 7^{401} equiv 3 + 7 = 10 pmod{500}$.
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
$endgroup$
Use Euler’s Theorem: If $gcd(a,m) = 1$ then $a^{phi(m)} equiv pmod{m}$, where $phi$ is Euler's totient function. Hence, if $gcd(a,m) = 1$ then $x equiv y pmod{phi(m)} Rightarrow a^x equiv a^y pmod{m}$. We have $phi(500) = 200$, and $401 equiv 1 pmod{200}$, hence $3^{401} + 7^{401} equiv 3 + 7 = 10 pmod{500}$.
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
edited Dec 1 '18 at 23:56
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
answered Dec 1 '18 at 23:50
mlerma54mlerma54
1,162138
1,162138
add a comment |
add a comment |
1
$begingroup$
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.
$endgroup$
– Carl Mummert
Dec 6 '18 at 17:50