Solving modular exponentiation
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Calculate : $(8^{2^{6^{4^{2^{5^{8^9}}}}}}) (mod 10000)$
But, the problem is that $8$ and $10000$ are not co-prime.
Moreover, the goal is to use Euler's theorem (modified?) to solve this. Any help is appreciated.
number-theory modular-arithmetic exponentiation totient-function
edited Nov 23 at 7:20
M.Mass
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asked Nov 23 at 7:06
Aizen
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up vote
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favorite
Calculate : $(8^{2^{6^{4^{2^{5^{8^9}}}}}}) (mod 10000)$
But, the problem is that $8$ and $10000$ are not co-prime.
Moreover, the goal is to use Euler's theorem (modified?) to solve this. Any help is appreciated.
number-theory modular-arithmetic exponentiation totient-function
edited Nov 23 at 7:20
M.Mass
1,8493923
asked Nov 23 at 7:06
Aizen
136
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Calculate : $(8^{2^{6^{4^{2^{5^{8^9}}}}}}) (mod 10000)$
But, the problem is that $8$ and $10000$ are not co-prime.
Moreover, the goal is to use Euler's theorem (modified?) to solve this. Any help is appreciated.
number-theory modular-arithmetic exponentiation totient-function
edited Nov 23 at 7:20
M.Mass
1,8493923
asked Nov 23 at 7:06
Aizen
136
Calculate : $(8^{2^{6^{4^{2^{5^{8^9}}}}}}) (mod 10000)$
But, the problem is that $8$ and $10000$ are not co-prime.
Moreover, the goal is to use Euler's theorem (modified?) to solve this. Any help is appreciated.
number-theory modular-arithmetic exponentiation totient-function
number-theory modular-arithmetic exponentiation totient-function
edited Nov 23 at 7:20
M.Mass
1,8493923
asked Nov 23 at 7:06
Aizen
136
edited Nov 23 at 7:20
M.Mass
1,8493923
asked Nov 23 at 7:06
Aizen
136
edited Nov 23 at 7:20
M.Mass
1,8493923
edited Nov 23 at 7:20
M.Mass
1,8493923
edited Nov 23 at 7:20
M.Mass
1,8493923
1,8493923
asked Nov 23 at 7:06
Aizen
136
asked Nov 23 at 7:06
Aizen
136
asked Nov 23 at 7:06
Aizen
136
136
add a comment |
add a comment |
1 Answer
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HINT: since they are not coprime, what can you say about the powers of $2$ involved here? And then work with the powers of $5$ and combine results (this is related to the Chinese Remainder Theorem, but you don't need to know that actually to solve the problem).
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
HINT: since they are not coprime, what can you say about the powers of $2$ involved here? And then work with the powers of $5$ and combine results (this is related to the Chinese Remainder Theorem, but you don't need to know that actually to solve the problem).
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
add a comment |
up vote
2
down vote
HINT: since they are not coprime, what can you say about the powers of $2$ involved here? And then work with the powers of $5$ and combine results (this is related to the Chinese Remainder Theorem, but you don't need to know that actually to solve the problem).
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT: since they are not coprime, what can you say about the powers of $2$ involved here? And then work with the powers of $5$ and combine results (this is related to the Chinese Remainder Theorem, but you don't need to know that actually to solve the problem).
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
HINT: since they are not coprime, what can you say about the powers of $2$ involved here? And then work with the powers of $5$ and combine results (this is related to the Chinese Remainder Theorem, but you don't need to know that actually to solve the problem).
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
answered Nov 23 at 7:15
Mark Bennet
80.1k981179
80.1k981179
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
add a comment |
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
I could use a better explanation. Thanks for the comment, but i need a bit more explanation.
– Aizen
Nov 25 at 4:37
add a comment |
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