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Finding power of $2$ and $3$ in $19^88-1$ [closed]

-1

1

$begingroup$

Find the exponent of 2 and 3 in $$19^88 -1$$.
Do we need to think with Fermat theorem?

elementary-number-theory

share|cite|improve this question

edited Mar 24 at 6:30

David G. Stork

12.1k41836

asked Mar 24 at 6:20

mavericmaveric

89612

$endgroup$

closed as off-topic by Eevee Trainer, Jyrki Lahtonen, John Omielan, Abcd, mrtaurho Apr 1 at 6:47

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." – Eevee Trainer, Jyrki Lahtonen, John Omielan, Abcd, mrtaurho

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $2^5$ and $3^2$
  $endgroup$
  – David G. Stork
  Mar 24 at 6:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$(1+18)^88-1equivbinom88118pmod3^3$$
  $endgroup$
  – lab bhattacharjee
  Mar 24 at 6:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please show us your current work, so we can build on it?
  $endgroup$
  – BadAtGeometry
  Mar 24 at 6:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i could not attempt.
  $endgroup$
  – maveric
  Mar 24 at 7:03
  

  
  
  
  

add a comment | 

-1

1

$begingroup$

Find the exponent of 2 and 3 in $$19^88 -1$$.
Do we need to think with Fermat theorem?

elementary-number-theory

share|cite|improve this question

edited Mar 24 at 6:30

David G. Stork

12.1k41836

asked Mar 24 at 6:20

mavericmaveric

89612

$endgroup$

closed as off-topic by Eevee Trainer, Jyrki Lahtonen, John Omielan, Abcd, mrtaurho Apr 1 at 6:47

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." – Eevee Trainer, Jyrki Lahtonen, John Omielan, Abcd, mrtaurho

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $2^5$ and $3^2$
  $endgroup$
  – David G. Stork
  Mar 24 at 6:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$(1+18)^88-1equivbinom88118pmod3^3$$
  $endgroup$
  – lab bhattacharjee
  Mar 24 at 6:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please show us your current work, so we can build on it?
  $endgroup$
  – BadAtGeometry
  Mar 24 at 6:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i could not attempt.
  $endgroup$
  – maveric
  Mar 24 at 7:03
  

  
  
  
  

add a comment | 

$begingroup$

Find the exponent of 2 and 3 in $$19^88 -1$$.
Do we need to think with Fermat theorem?

elementary-number-theory

share|cite|improve this question

edited Mar 24 at 6:30

David G. Stork

12.1k41836

asked Mar 24 at 6:20

mavericmaveric

89612

$endgroup$

share|cite|improve this question

edited Mar 24 at 6:30

David G. Stork

12.1k41836

asked Mar 24 at 6:20

mavericmaveric

89612

share|cite|improve this question

edited Mar 24 at 6:30

David G. Stork

12.1k41836

asked Mar 24 at 6:20

mavericmaveric

89612

edited Mar 24 at 6:30

David G. Stork

12.1k41836

edited Mar 24 at 6:30

David G. Stork

12.1k41836

asked Mar 24 at 6:20

mavericmaveric

89612

asked Mar 24 at 6:20

mavericmaveric

89612

2 Answers
2

active

oldest

votes

2

$begingroup$

$19^88-1=(19^11-1)(19^11+1)(19^22+1)(19^44+1)$

$19^11-1=(19-1)(19^10+19^9+ . . . +1)=2times 3^2times k$

$19^11+1=(19+1)(19^10-19^9+ . . . +1)=2^2times 5times k_1$

$19^22+1=2 k_2$

$19^44+1=2 k_3$

$19^88-1=(2times 3^2 k)times (2^2times 5 k_1)times(2 k_2)times (2 k_2)=2^5times 3^2times 5. k.k_1.k_2 $

share|cite|improve this answer

answered Mar 24 at 9:47

siroussirous

1,6981514

$endgroup$

add a comment | 

-1

$begingroup$

If $a=2^mbpm1$ where $b$ is odd

$a^2=1pm2^m+1b+b^22^2m$

So, the highest power of $2$ in $a^2-1$ will be $2^m+1$ for $2mge m+1iff mge1$

So, the highest power of $2$ in $(2^2cdot5-1)^8-1$ will be $2^2+3$

So we can set $19^8=1+32c$ where $c$ is odd

$19^88=(1+32c)^11equiv1+32c(11)pmod32^2$

share|cite|improve this answer

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can we do without mod?
  $endgroup$
  – maveric
  Mar 24 at 7:03
  

  
  
  
  

add a comment | 

2

$begingroup$

$19^88-1=(19^11-1)(19^11+1)(19^22+1)(19^44+1)$

$19^11-1=(19-1)(19^10+19^9+ . . . +1)=2times 3^2times k$

$19^11+1=(19+1)(19^10-19^9+ . . . +1)=2^2times 5times k_1$

$19^22+1=2 k_2$

$19^44+1=2 k_3$

$19^88-1=(2times 3^2 k)times (2^2times 5 k_1)times(2 k_2)times (2 k_2)=2^5times 3^2times 5. k.k_1.k_2 $

share|cite|improve this answer

answered Mar 24 at 9:47

siroussirous

1,6981514

$endgroup$

add a comment | 

2

$begingroup$

$19^88-1=(19^11-1)(19^11+1)(19^22+1)(19^44+1)$

$19^11-1=(19-1)(19^10+19^9+ . . . +1)=2times 3^2times k$

$19^11+1=(19+1)(19^10-19^9+ . . . +1)=2^2times 5times k_1$

$19^22+1=2 k_2$

$19^44+1=2 k_3$

$19^88-1=(2times 3^2 k)times (2^2times 5 k_1)times(2 k_2)times (2 k_2)=2^5times 3^2times 5. k.k_1.k_2 $

share|cite|improve this answer

answered Mar 24 at 9:47

siroussirous

1,6981514

$endgroup$

add a comment | 

$begingroup$

$19^88-1=(19^11-1)(19^11+1)(19^22+1)(19^44+1)$

$19^11-1=(19-1)(19^10+19^9+ . . . +1)=2times 3^2times k$

$19^11+1=(19+1)(19^10-19^9+ . . . +1)=2^2times 5times k_1$

$19^22+1=2 k_2$

$19^44+1=2 k_3$

$19^88-1=(2times 3^2 k)times (2^2times 5 k_1)times(2 k_2)times (2 k_2)=2^5times 3^2times 5. k.k_1.k_2 $

share|cite|improve this answer

answered Mar 24 at 9:47

siroussirous

1,6981514

$endgroup$

share|cite|improve this answer

answered Mar 24 at 9:47

siroussirous

1,6981514

answered Mar 24 at 9:47

siroussirous

1,6981514

answered Mar 24 at 9:47

siroussirous

1,6981514

-1

$begingroup$

If $a=2^mbpm1$ where $b$ is odd

$a^2=1pm2^m+1b+b^22^2m$

So, the highest power of $2$ in $a^2-1$ will be $2^m+1$ for $2mge m+1iff mge1$

So, the highest power of $2$ in $(2^2cdot5-1)^8-1$ will be $2^2+3$

So we can set $19^8=1+32c$ where $c$ is odd

$19^88=(1+32c)^11equiv1+32c(11)pmod32^2$

share|cite|improve this answer

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can we do without mod?
  $endgroup$
  – maveric
  Mar 24 at 7:03
  

  
  
  
  

add a comment | 

-1

$begingroup$

If $a=2^mbpm1$ where $b$ is odd

$a^2=1pm2^m+1b+b^22^2m$

So, the highest power of $2$ in $a^2-1$ will be $2^m+1$ for $2mge m+1iff mge1$

So, the highest power of $2$ in $(2^2cdot5-1)^8-1$ will be $2^2+3$

So we can set $19^8=1+32c$ where $c$ is odd

$19^88=(1+32c)^11equiv1+32c(11)pmod32^2$

share|cite|improve this answer

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can we do without mod?
  $endgroup$
  – maveric
  Mar 24 at 7:03
  

  
  
  
  

add a comment | 

$begingroup$

If $a=2^mbpm1$ where $b$ is odd

$a^2=1pm2^m+1b+b^22^2m$

So, the highest power of $2$ in $a^2-1$ will be $2^m+1$ for $2mge m+1iff mge1$

So, the highest power of $2$ in $(2^2cdot5-1)^8-1$ will be $2^2+3$

So we can set $19^8=1+32c$ where $c$ is odd

$19^88=(1+32c)^11equiv1+32c(11)pmod32^2$

share|cite|improve this answer

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

share|cite|improve this answer

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279

answered Mar 24 at 6:43

lab bhattacharjeelab bhattacharjee

228k15159279