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Prove that $frac(72!)(36!)^2-1$ is divisible by 73 [closed]

1

1

$begingroup$

Prove that $frac(72!)(36!)^2-1$ is divisible by 73.

My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.

modular-arithmetic divisibility natural-numbers

share|cite|improve this question

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

$endgroup$

closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRL Apr 4 at 16:20

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." – user21820, José Carlos Santos, Adrian Keister, K.Power, RRL

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $73-aequiv-apmod73$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 31 at 6:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $binom72kequiv(-1)^kpmod73$
  $endgroup$
  – robjohn♦
  Mar 31 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Special case of this.
  $endgroup$
  – Bill Dubuque
  Apr 1 at 3:04
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Prove that $frac(72!)(36!)^2-1$ is divisible by 73.

My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.

modular-arithmetic divisibility natural-numbers

share|cite|improve this question

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

$endgroup$

closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRL Apr 4 at 16:20

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." – user21820, José Carlos Santos, Adrian Keister, K.Power, RRL

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $73-aequiv-apmod73$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 31 at 6:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $binom72kequiv(-1)^kpmod73$
  $endgroup$
  – robjohn♦
  Mar 31 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Special case of this.
  $endgroup$
  – Bill Dubuque
  Apr 1 at 3:04
  

  
  
  
  

add a comment | 

$begingroup$

Prove that $frac(72!)(36!)^2-1$ is divisible by 73.

My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.

modular-arithmetic divisibility natural-numbers

share|cite|improve this question

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

$endgroup$

share|cite|improve this question

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

share|cite|improve this question

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

edited Mar 31 at 9:05

Michael Rozenberg

110k1896201

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

asked Mar 31 at 6:42

Samar Imam ZaidiSamar Imam Zaidi

1,6221520

2 Answers
2

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3

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Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$

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answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

add a comment | 

2

$begingroup$

Hint: Prove that $$72! equiv (36!)^2 pmod73$$

Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$

Since 73 is prime....

share|cite|improve this answer

answered Mar 31 at 6:50

N. S.N. S.

105k7115210

$endgroup$

add a comment | 

3

$begingroup$

Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$

share|cite|improve this answer

answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

add a comment | 

3

$begingroup$

Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$

share|cite|improve this answer

answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

add a comment | 

$begingroup$

Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$

share|cite|improve this answer

answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

share|cite|improve this answer

answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 31 at 8:03

Michael RozenbergMichael Rozenberg

110k1896201

2

$begingroup$

Hint: Prove that $$72! equiv (36!)^2 pmod73$$

Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$

Since 73 is prime....

share|cite|improve this answer

answered Mar 31 at 6:50

N. S.N. S.

105k7115210

$endgroup$

add a comment | 

2

$begingroup$

Hint: Prove that $$72! equiv (36!)^2 pmod73$$

Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$

Since 73 is prime....

share|cite|improve this answer

answered Mar 31 at 6:50

N. S.N. S.

105k7115210

$endgroup$

add a comment | 

$begingroup$

Hint: Prove that $$72! equiv (36!)^2 pmod73$$

Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$

Since 73 is prime....

share|cite|improve this answer

answered Mar 31 at 6:50

N. S.N. S.

105k7115210

$endgroup$

share|cite|improve this answer

answered Mar 31 at 6:50

N. S.N. S.

105k7115210

answered Mar 31 at 6:50

N. S.N. S.

105k7115210

answered Mar 31 at 6:50

N. S.N. S.

105k7115210