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Find $438^{87493} equiv ~? pmod{11}$ [closed]

-1

1

$begingroup$

How to find the value of '?'

a mod m = b mod m , will this formula be used?

I am taking discrete maths course for CS. And this question is from one of its chapter

modular-arithmetic

share|cite|improve this question

edited Mar 21 at 6:51

rdr2

asked Mar 21 at 6:32

rdr2rdr2

12

$endgroup$

closed as off-topic by José Carlos Santos, Saad, Cesareo, Arnaud D., Javi Mar 21 at 12:46

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." – José Carlos Santos, Saad, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What is $438 (mod 11)$? And what is the order of $(Bbb Z / 11 Bbb Z)^*$?
  $endgroup$
  – Robert Shore
  Mar 21 at 6:39
  
  
  

  
  
  
  


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  $begingroup$
  438 mod 11 would be 9 And I don't know what you mean by order of (Z/11Z)
  $endgroup$
  – rdr2
  Mar 21 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For us to be able to help you effectively, you need to provide more background. Why is this problem of interest to you? How have you tried to solve it?
  $endgroup$
  – Robert Shore
  Mar 21 at 6:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am taking discrete maths course for CS. And this question is from one of its chapter
  $endgroup$
  – rdr2
  Mar 21 at 6:47
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of How do I compute $a^b,bmod c$ by hand?
  $endgroup$
  – Arnaud D.
  Mar 21 at 10:47
  

  
  
  
  

 | 
show 7 more comments

-1

1

$begingroup$

How to find the value of '?'

a mod m = b mod m , will this formula be used?

I am taking discrete maths course for CS. And this question is from one of its chapter

modular-arithmetic

share|cite|improve this question

edited Mar 21 at 6:51

rdr2

asked Mar 21 at 6:32

rdr2rdr2

12

$endgroup$

closed as off-topic by José Carlos Santos, Saad, Cesareo, Arnaud D., Javi Mar 21 at 12:46

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." – José Carlos Santos, Saad, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What is $438 (mod 11)$? And what is the order of $(Bbb Z / 11 Bbb Z)^*$?
  $endgroup$
  – Robert Shore
  Mar 21 at 6:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  438 mod 11 would be 9 And I don't know what you mean by order of (Z/11Z)
  $endgroup$
  – rdr2
  Mar 21 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For us to be able to help you effectively, you need to provide more background. Why is this problem of interest to you? How have you tried to solve it?
  $endgroup$
  – Robert Shore
  Mar 21 at 6:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am taking discrete maths course for CS. And this question is from one of its chapter
  $endgroup$
  – rdr2
  Mar 21 at 6:47
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of How do I compute $a^b,bmod c$ by hand?
  $endgroup$
  – Arnaud D.
  Mar 21 at 10:47
  

  
  
  
  

 | 
show 7 more comments

$begingroup$

How to find the value of '?'

a mod m = b mod m , will this formula be used?

I am taking discrete maths course for CS. And this question is from one of its chapter

modular-arithmetic

share|cite|improve this question

edited Mar 21 at 6:51

rdr2

asked Mar 21 at 6:32

rdr2rdr2

12

$endgroup$

share|cite|improve this question

edited Mar 21 at 6:51

rdr2

asked Mar 21 at 6:32

rdr2rdr2

12

share|cite|improve this question

edited Mar 21 at 6:51

rdr2

asked Mar 21 at 6:32

rdr2rdr2

12

asked Mar 21 at 6:32

rdr2rdr2

12

asked Mar 21 at 6:32

rdr2rdr2

12

1 Answer
1

active

oldest

votes

2

$begingroup$

Observe that $text {gcd} (11,438) = 1.$ So by Euler's theorem $438^{10} equiv 1 (text {mod} 11).$ So $438^{87490} equiv 1 (text {mod} 11).$ Also $438 equiv -2 (text {mod} 11) implies 438^3 equiv -8 equiv 3 (text {mod} 11).$ Therefore $$438^{87493} equiv 3 (text {mod} 11).$$

If you don't know Euler's theorem or Fermat's little theorem then observe from my above calculation that $438^3 equiv 3 (text {mod} 11).$ So $438^6 equiv 9 equiv -2 (text {mod} 11) implies 438^{30} equiv (-2)^5 equiv -32 equiv 1 (text {mod} 11).$

Observe that $87493 = 87480 + 13.$ Since $30 mid 87480$ so $438^{87480} equiv 1 (text {mod} 11).$ Since $438^6 equiv -2 (text {mod} 11)$ so $438^{12} equiv 4 (text {mod} 11).$ Again $438 equiv -2 (text {mod} 11).$ So $438^{13} equiv 438^{12} cdot 438 equiv 4 cdot (-2) equiv -8 equiv 3 (text {mod} 11).$ Thus we get

$(1)$ $438^{87480} equiv 1 (text {mod} 11).$

$(2)$ $438^{13} equiv 3 (text {mod} 11).$

Therefore what is $438^{87493} equiv ~? (text {mod} 11)$?

$$438^{87493} equiv 438^{87480+13} equiv 438^{87490} cdot 438^{13} equiv 1 cdot 3 equiv 3 (text {mod} 11).$$

share|cite|improve this answer

edited Mar 21 at 8:03

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622

$endgroup$

add a comment | 

2

$begingroup$

Observe that $text {gcd} (11,438) = 1.$ So by Euler's theorem $438^{10} equiv 1 (text {mod} 11).$ So $438^{87490} equiv 1 (text {mod} 11).$ Also $438 equiv -2 (text {mod} 11) implies 438^3 equiv -8 equiv 3 (text {mod} 11).$ Therefore $$438^{87493} equiv 3 (text {mod} 11).$$

If you don't know Euler's theorem or Fermat's little theorem then observe from my above calculation that $438^3 equiv 3 (text {mod} 11).$ So $438^6 equiv 9 equiv -2 (text {mod} 11) implies 438^{30} equiv (-2)^5 equiv -32 equiv 1 (text {mod} 11).$

Observe that $87493 = 87480 + 13.$ Since $30 mid 87480$ so $438^{87480} equiv 1 (text {mod} 11).$ Since $438^6 equiv -2 (text {mod} 11)$ so $438^{12} equiv 4 (text {mod} 11).$ Again $438 equiv -2 (text {mod} 11).$ So $438^{13} equiv 438^{12} cdot 438 equiv 4 cdot (-2) equiv -8 equiv 3 (text {mod} 11).$ Thus we get

$(1)$ $438^{87480} equiv 1 (text {mod} 11).$

$(2)$ $438^{13} equiv 3 (text {mod} 11).$

Therefore what is $438^{87493} equiv ~? (text {mod} 11)$?

$$438^{87493} equiv 438^{87480+13} equiv 438^{87490} cdot 438^{13} equiv 1 cdot 3 equiv 3 (text {mod} 11).$$

share|cite|improve this answer

edited Mar 21 at 8:03

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622

$endgroup$

add a comment | 

2

$begingroup$

Observe that $text {gcd} (11,438) = 1.$ So by Euler's theorem $438^{10} equiv 1 (text {mod} 11).$ So $438^{87490} equiv 1 (text {mod} 11).$ Also $438 equiv -2 (text {mod} 11) implies 438^3 equiv -8 equiv 3 (text {mod} 11).$ Therefore $$438^{87493} equiv 3 (text {mod} 11).$$

If you don't know Euler's theorem or Fermat's little theorem then observe from my above calculation that $438^3 equiv 3 (text {mod} 11).$ So $438^6 equiv 9 equiv -2 (text {mod} 11) implies 438^{30} equiv (-2)^5 equiv -32 equiv 1 (text {mod} 11).$

Observe that $87493 = 87480 + 13.$ Since $30 mid 87480$ so $438^{87480} equiv 1 (text {mod} 11).$ Since $438^6 equiv -2 (text {mod} 11)$ so $438^{12} equiv 4 (text {mod} 11).$ Again $438 equiv -2 (text {mod} 11).$ So $438^{13} equiv 438^{12} cdot 438 equiv 4 cdot (-2) equiv -8 equiv 3 (text {mod} 11).$ Thus we get

$(1)$ $438^{87480} equiv 1 (text {mod} 11).$

$(2)$ $438^{13} equiv 3 (text {mod} 11).$

Therefore what is $438^{87493} equiv ~? (text {mod} 11)$?

$$438^{87493} equiv 438^{87480+13} equiv 438^{87490} cdot 438^{13} equiv 1 cdot 3 equiv 3 (text {mod} 11).$$

share|cite|improve this answer

edited Mar 21 at 8:03

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622

$endgroup$

add a comment | 

$begingroup$

Observe that $text {gcd} (11,438) = 1.$ So by Euler's theorem $438^{10} equiv 1 (text {mod} 11).$ So $438^{87490} equiv 1 (text {mod} 11).$ Also $438 equiv -2 (text {mod} 11) implies 438^3 equiv -8 equiv 3 (text {mod} 11).$ Therefore $$438^{87493} equiv 3 (text {mod} 11).$$

If you don't know Euler's theorem or Fermat's little theorem then observe from my above calculation that $438^3 equiv 3 (text {mod} 11).$ So $438^6 equiv 9 equiv -2 (text {mod} 11) implies 438^{30} equiv (-2)^5 equiv -32 equiv 1 (text {mod} 11).$

Observe that $87493 = 87480 + 13.$ Since $30 mid 87480$ so $438^{87480} equiv 1 (text {mod} 11).$ Since $438^6 equiv -2 (text {mod} 11)$ so $438^{12} equiv 4 (text {mod} 11).$ Again $438 equiv -2 (text {mod} 11).$ So $438^{13} equiv 438^{12} cdot 438 equiv 4 cdot (-2) equiv -8 equiv 3 (text {mod} 11).$ Thus we get

$(1)$ $438^{87480} equiv 1 (text {mod} 11).$

$(2)$ $438^{13} equiv 3 (text {mod} 11).$

Therefore what is $438^{87493} equiv ~? (text {mod} 11)$?

$$438^{87493} equiv 438^{87480+13} equiv 438^{87490} cdot 438^{13} equiv 1 cdot 3 equiv 3 (text {mod} 11).$$

share|cite|improve this answer

edited Mar 21 at 8:03

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622

$endgroup$

share|cite|improve this answer

edited Mar 21 at 8:03

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622

answered Mar 21 at 6:45

Dbchatto67Dbchatto67

2,753622