Chinese Remainder Theorem - solving a modulo with big numbers












1












$begingroup$


I have the calculation: $2^{31}pmod {2925}$



It's for university and we should solve it like:




  1. make prime partition

  2. $2^{31}$ mod all prime partitions

  3. Solve with Chinese Remainder Theorem.


I started with $2925 = 3 cdot 3 cdot 5 cdot 5 cdot 13$ , and found out that:
$$2^{31} equiv 2 pmod{3}$$
$$2^{31} equiv 3 pmod{5}$$
$$2^{31} equiv 11 pmod{13}$$
I made:
$$x equiv 2 pmod3$$
$$x equiv 3 pmod5$$
$$x equiv 11 pmod{13}$$



Then I tried CRT and got $x = -1237 + 195k$



If you simply calculate $2^{31}pmod{ 2925}$ you get $1298$, which is in fact $-1237 + 195 cdot 13$.



I don't know how to find out the $13$.



Any help appreciated.



EDIT:



SOLVED!
I took $3$ instead of $9$ and $5$ instead of $25$ after prime partition. For more infos please see comments. Thanks!






chinese-remainder-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    For CRT you need to use moduli $,9,25,13,,$ not $,3,5,13,$
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:13












  • $begingroup$
    Ah, so I need to multiply the same primes.. like if I had 3*3*3 instead of 3*3 I had to take 27?
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:16










  • $begingroup$
    You want the lcm of the moduli to be $2925$ in order to get the result mod $2925$, and you want them pairwise coprime so you can apply CRT.
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:19












  • $begingroup$
    Okay I see & I'will try. Thank you !
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:23










  • $begingroup$
    Worked out perfectly, thank you :)
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:44
















1












$begingroup$


I have the calculation: $2^{31}pmod {2925}$



It's for university and we should solve it like:




  1. make prime partition

  2. $2^{31}$ mod all prime partitions

  3. Solve with Chinese Remainder Theorem.


I started with $2925 = 3 cdot 3 cdot 5 cdot 5 cdot 13$ , and found out that:
$$2^{31} equiv 2 pmod{3}$$
$$2^{31} equiv 3 pmod{5}$$
$$2^{31} equiv 11 pmod{13}$$
I made:
$$x equiv 2 pmod3$$
$$x equiv 3 pmod5$$
$$x equiv 11 pmod{13}$$



Then I tried CRT and got $x = -1237 + 195k$



If you simply calculate $2^{31}pmod{ 2925}$ you get $1298$, which is in fact $-1237 + 195 cdot 13$.



I don't know how to find out the $13$.



Any help appreciated.



EDIT:



SOLVED!
I took $3$ instead of $9$ and $5$ instead of $25$ after prime partition. For more infos please see comments. Thanks!






chinese-remainder-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    For CRT you need to use moduli $,9,25,13,,$ not $,3,5,13,$
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:13












  • $begingroup$
    Ah, so I need to multiply the same primes.. like if I had 3*3*3 instead of 3*3 I had to take 27?
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:16










  • $begingroup$
    You want the lcm of the moduli to be $2925$ in order to get the result mod $2925$, and you want them pairwise coprime so you can apply CRT.
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:19












  • $begingroup$
    Okay I see & I'will try. Thank you !
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:23










  • $begingroup$
    Worked out perfectly, thank you :)
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:44














1












1








1


0



$begingroup$


I have the calculation: $2^{31}pmod {2925}$



It's for university and we should solve it like:




  1. make prime partition

  2. $2^{31}$ mod all prime partitions

  3. Solve with Chinese Remainder Theorem.


I started with $2925 = 3 cdot 3 cdot 5 cdot 5 cdot 13$ , and found out that:
$$2^{31} equiv 2 pmod{3}$$
$$2^{31} equiv 3 pmod{5}$$
$$2^{31} equiv 11 pmod{13}$$
I made:
$$x equiv 2 pmod3$$
$$x equiv 3 pmod5$$
$$x equiv 11 pmod{13}$$



Then I tried CRT and got $x = -1237 + 195k$



If you simply calculate $2^{31}pmod{ 2925}$ you get $1298$, which is in fact $-1237 + 195 cdot 13$.



I don't know how to find out the $13$.



Any help appreciated.



EDIT:



SOLVED!
I took $3$ instead of $9$ and $5$ instead of $25$ after prime partition. For more infos please see comments. Thanks!






chinese-remainder-theorem







share|cite|improve this question











$endgroup$




I have the calculation: $2^{31}pmod {2925}$



It's for university and we should solve it like:




  1. make prime partition

  2. $2^{31}$ mod all prime partitions

  3. Solve with Chinese Remainder Theorem.


I started with $2925 = 3 cdot 3 cdot 5 cdot 5 cdot 13$ , and found out that:
$$2^{31} equiv 2 pmod{3}$$
$$2^{31} equiv 3 pmod{5}$$
$$2^{31} equiv 11 pmod{13}$$
I made:
$$x equiv 2 pmod3$$
$$x equiv 3 pmod5$$
$$x equiv 11 pmod{13}$$



Then I tried CRT and got $x = -1237 + 195k$



If you simply calculate $2^{31}pmod{ 2925}$ you get $1298$, which is in fact $-1237 + 195 cdot 13$.



I don't know how to find out the $13$.



Any help appreciated.



EDIT:



SOLVED!
I took $3$ instead of $9$ and $5$ instead of $25$ after prime partition. For more infos please see comments. Thanks!






chinese-remainder-theorem




chinese-remainder-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 20 '18 at 12:30









JayTuma

1,326218




1,326218










asked Apr 24 '15 at 19:02









SomebodySomebody

1098




1098












  • $begingroup$
    For CRT you need to use moduli $,9,25,13,,$ not $,3,5,13,$
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:13












  • $begingroup$
    Ah, so I need to multiply the same primes.. like if I had 3*3*3 instead of 3*3 I had to take 27?
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:16










  • $begingroup$
    You want the lcm of the moduli to be $2925$ in order to get the result mod $2925$, and you want them pairwise coprime so you can apply CRT.
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:19












  • $begingroup$
    Okay I see & I'will try. Thank you !
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:23










  • $begingroup$
    Worked out perfectly, thank you :)
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:44


















  • $begingroup$
    For CRT you need to use moduli $,9,25,13,,$ not $,3,5,13,$
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:13












  • $begingroup$
    Ah, so I need to multiply the same primes.. like if I had 3*3*3 instead of 3*3 I had to take 27?
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:16










  • $begingroup$
    You want the lcm of the moduli to be $2925$ in order to get the result mod $2925$, and you want them pairwise coprime so you can apply CRT.
    $endgroup$
    – Bill Dubuque
    Apr 24 '15 at 19:19












  • $begingroup$
    Okay I see & I'will try. Thank you !
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:23










  • $begingroup$
    Worked out perfectly, thank you :)
    $endgroup$
    – Somebody
    Apr 24 '15 at 19:44
















$begingroup$
For CRT you need to use moduli $,9,25,13,,$ not $,3,5,13,$
$endgroup$
– Bill Dubuque
Apr 24 '15 at 19:13






$begingroup$
For CRT you need to use moduli $,9,25,13,,$ not $,3,5,13,$
$endgroup$
– Bill Dubuque
Apr 24 '15 at 19:13














$begingroup$
Ah, so I need to multiply the same primes.. like if I had 3*3*3 instead of 3*3 I had to take 27?
$endgroup$
– Somebody
Apr 24 '15 at 19:16




$begingroup$
Ah, so I need to multiply the same primes.. like if I had 3*3*3 instead of 3*3 I had to take 27?
$endgroup$
– Somebody
Apr 24 '15 at 19:16












$begingroup$
You want the lcm of the moduli to be $2925$ in order to get the result mod $2925$, and you want them pairwise coprime so you can apply CRT.
$endgroup$
– Bill Dubuque
Apr 24 '15 at 19:19






$begingroup$
You want the lcm of the moduli to be $2925$ in order to get the result mod $2925$, and you want them pairwise coprime so you can apply CRT.
$endgroup$
– Bill Dubuque
Apr 24 '15 at 19:19














$begingroup$
Okay I see & I'will try. Thank you !
$endgroup$
– Somebody
Apr 24 '15 at 19:23




$begingroup$
Okay I see & I'will try. Thank you !
$endgroup$
– Somebody
Apr 24 '15 at 19:23












$begingroup$
Worked out perfectly, thank you :)
$endgroup$
– Somebody
Apr 24 '15 at 19:44




$begingroup$
Worked out perfectly, thank you :)
$endgroup$
– Somebody
Apr 24 '15 at 19:44










1 Answer
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$begingroup$

For CRT you need to use moduli $9,25,13$ not $3,5,13$



You want the l.c.m. of the moduli to be $2925$ in order to get the result modulo $2925$, and you want them pairwise coprime so you can apply CRT.



Thanks to Bill Dubuque for the answer.






share|cite|improve this answer











$endgroup$













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    $begingroup$

    For CRT you need to use moduli $9,25,13$ not $3,5,13$



    You want the l.c.m. of the moduli to be $2925$ in order to get the result modulo $2925$, and you want them pairwise coprime so you can apply CRT.



    Thanks to Bill Dubuque for the answer.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      For CRT you need to use moduli $9,25,13$ not $3,5,13$



      You want the l.c.m. of the moduli to be $2925$ in order to get the result modulo $2925$, and you want them pairwise coprime so you can apply CRT.



      Thanks to Bill Dubuque for the answer.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        For CRT you need to use moduli $9,25,13$ not $3,5,13$



        You want the l.c.m. of the moduli to be $2925$ in order to get the result modulo $2925$, and you want them pairwise coprime so you can apply CRT.



        Thanks to Bill Dubuque for the answer.






        share|cite|improve this answer











        $endgroup$



        For CRT you need to use moduli $9,25,13$ not $3,5,13$



        You want the l.c.m. of the moduli to be $2925$ in order to get the result modulo $2925$, and you want them pairwise coprime so you can apply CRT.



        Thanks to Bill Dubuque for the answer.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 20 '18 at 12:25









        JayTuma

        1,326218




        1,326218










        answered Apr 24 '15 at 19:48









        SomebodySomebody

        1098




        1098






























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