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Nested exponent modulus, $2^{2^{517}} pmod {23}$

What is the remainder when $2^{2^{517}}$ divided by $23$?Modulo Problem, Fermat's little theoremDetermine the smallest integer $k$ such that $4^k equiv 1 pmod{19}$.Solving a system of modular equatiosHow to solve this modular equation? $x^{19} equiv 36 mod 97$.Why is $a^{5} equiv apmod 5$ for any positive integer?Simultaneusly solving $2x equiv 11 pmod{15}$ and $3x equiv 6 pmod 8$How to evaluate $1234^{1234} pmod{5379}$?Modular arithmetic system $x equiv 2 pmod{7}$ and $x equiv -5 pmod{22}$Show that $a^{(p-1)(q-1)} equiv 1 pmod{n}$Prove or disprove $a^{10}=b^{10} pmod{10alpha}$

0

1

$begingroup$

I have had a go at this for a while now and can't seem to get anywhere.

$$2^{(2^{517})} pmod {23}$$

I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.

Here is my attempt:

$$2^{(2^{517} !!mod {22})} equiv 2^{(2^{517})} pmod {23}$$
Then proceed to do Chinese remainder theorem to find the exponent?

number-theory modular-arithmetic

share|cite|improve this question

edited Mar 13 at 4:15

Rócherz

2,9863821

asked Mar 14 '16 at 3:38

softiesoftie

32

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Take $2^{517}$ mod $22$ as your new exponent
  $endgroup$
  – TokenToucan
  Mar 14 '16 at 3:44
  
  
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I have had a go at this for a while now and can't seem to get anywhere.

$$2^{(2^{517})} pmod {23}$$

I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.

Here is my attempt:

$$2^{(2^{517} !!mod {22})} equiv 2^{(2^{517})} pmod {23}$$
Then proceed to do Chinese remainder theorem to find the exponent?

number-theory modular-arithmetic

share|cite|improve this question

edited Mar 13 at 4:15

Rócherz

2,9863821

asked Mar 14 '16 at 3:38

softiesoftie

32

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Take $2^{517}$ mod $22$ as your new exponent
  $endgroup$
  – TokenToucan
  Mar 14 '16 at 3:44
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I have had a go at this for a while now and can't seem to get anywhere.

$$2^{(2^{517})} pmod {23}$$

I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.

Here is my attempt:

$$2^{(2^{517} !!mod {22})} equiv 2^{(2^{517})} pmod {23}$$
Then proceed to do Chinese remainder theorem to find the exponent?

number-theory modular-arithmetic

share|cite|improve this question

edited Mar 13 at 4:15

Rócherz

2,9863821

asked Mar 14 '16 at 3:38

softiesoftie

32

$endgroup$

share|cite|improve this question

edited Mar 13 at 4:15

Rócherz

2,9863821

asked Mar 14 '16 at 3:38

softiesoftie

32

share|cite|improve this question

edited Mar 13 at 4:15

Rócherz

2,9863821

asked Mar 14 '16 at 3:38

softiesoftie

32

edited Mar 13 at 4:15

Rócherz

2,9863821

edited Mar 13 at 4:15

Rócherz

2,9863821

asked Mar 14 '16 at 3:38

softiesoftie

32

asked Mar 14 '16 at 3:38

softiesoftie

32

1 Answer
1

active

oldest

votes

2

$begingroup$

Note that $2^{11}=2048equiv1pmod{23}$

Next note that: $2^{10}=1024equiv1pmod{11}$

So $2^{2^{517}}equiv2^{2^7}pmod{23}$ as $2^{517}equiv2^7pmod{11}$

Hence $2^{2^{517}}equiv2^{128}pmod{23}equiv2^7pmod{23}=128equiv13pmod{23}$

share|cite|improve this answer

edited Mar 14 '16 at 9:45

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is there a reason for the switch from 517 to 57?
  $endgroup$
  – softie
  Mar 14 '16 at 4:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A typo. I accidiently missed the 1. It doesn't change the working. Fixed now.
  $endgroup$
  – Ian Miller
  Mar 14 '16 at 9:45
  

  
  
  
  

add a comment | 

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2

$begingroup$

Note that $2^{11}=2048equiv1pmod{23}$

Next note that: $2^{10}=1024equiv1pmod{11}$

So $2^{2^{517}}equiv2^{2^7}pmod{23}$ as $2^{517}equiv2^7pmod{11}$

Hence $2^{2^{517}}equiv2^{128}pmod{23}equiv2^7pmod{23}=128equiv13pmod{23}$

share|cite|improve this answer

edited Mar 14 '16 at 9:45

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is there a reason for the switch from 517 to 57?
  $endgroup$
  – softie
  Mar 14 '16 at 4:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A typo. I accidiently missed the 1. It doesn't change the working. Fixed now.
  $endgroup$
  – Ian Miller
  Mar 14 '16 at 9:45
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that $2^{11}=2048equiv1pmod{23}$

Next note that: $2^{10}=1024equiv1pmod{11}$

So $2^{2^{517}}equiv2^{2^7}pmod{23}$ as $2^{517}equiv2^7pmod{11}$

Hence $2^{2^{517}}equiv2^{128}pmod{23}equiv2^7pmod{23}=128equiv13pmod{23}$

share|cite|improve this answer

edited Mar 14 '16 at 9:45

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is there a reason for the switch from 517 to 57?
  $endgroup$
  – softie
  Mar 14 '16 at 4:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A typo. I accidiently missed the 1. It doesn't change the working. Fixed now.
  $endgroup$
  – Ian Miller
  Mar 14 '16 at 9:45
  

  
  
  
  

add a comment | 

$begingroup$

Note that $2^{11}=2048equiv1pmod{23}$

Next note that: $2^{10}=1024equiv1pmod{11}$

So $2^{2^{517}}equiv2^{2^7}pmod{23}$ as $2^{517}equiv2^7pmod{11}$

Hence $2^{2^{517}}equiv2^{128}pmod{23}equiv2^7pmod{23}=128equiv13pmod{23}$

share|cite|improve this answer

edited Mar 14 '16 at 9:45

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438

$endgroup$

share|cite|improve this answer

edited Mar 14 '16 at 9:45

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438

answered Mar 14 '16 at 3:48

Ian MillerIan Miller

10.5k11438