Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N??












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Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N? where a is an arbitrary integer, and N is a prime. e.g. i know 22! mod 23 ≡ 22 using Wilson's theorem. lets say i want to know 22! mod 24, is there a way i can do so without solving the factorial?






modular-arithmetic factorial







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  • $begingroup$
    I added an answer which discusses the general case.
    $endgroup$
    – Bill Dubuque
    Jan 8 at 20:21
















1












$begingroup$


Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N? where a is an arbitrary integer, and N is a prime. e.g. i know 22! mod 23 ≡ 22 using Wilson's theorem. lets say i want to know 22! mod 24, is there a way i can do so without solving the factorial?






modular-arithmetic factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    I added an answer which discusses the general case.
    $endgroup$
    – Bill Dubuque
    Jan 8 at 20:21














1












1








1





$begingroup$


Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N? where a is an arbitrary integer, and N is a prime. e.g. i know 22! mod 23 ≡ 22 using Wilson's theorem. lets say i want to know 22! mod 24, is there a way i can do so without solving the factorial?






modular-arithmetic factorial







share|cite|improve this question











$endgroup$




Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N? where a is an arbitrary integer, and N is a prime. e.g. i know 22! mod 23 ≡ 22 using Wilson's theorem. lets say i want to know 22! mod 24, is there a way i can do so without solving the factorial?






modular-arithmetic factorial




modular-arithmetic factorial






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edited Jan 8 at 15:04







Malik Hamza Murtaza

















asked Jan 8 at 14:57









Malik Hamza MurtazaMalik Hamza Murtaza

317




317












  • $begingroup$
    I added an answer which discusses the general case.
    $endgroup$
    – Bill Dubuque
    Jan 8 at 20:21


















  • $begingroup$
    I added an answer which discusses the general case.
    $endgroup$
    – Bill Dubuque
    Jan 8 at 20:21
















$begingroup$
I added an answer which discusses the general case.
$endgroup$
– Bill Dubuque
Jan 8 at 20:21




$begingroup$
I added an answer which discusses the general case.
$endgroup$
– Bill Dubuque
Jan 8 at 20:21










2 Answers
2






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oldest

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1












$begingroup$

If $,d := gcd(n,m)=1,$ then there is no relationship between $,a = xbmod n,$ and $, b = xbmod m,$ since CRT $,Rightarrow,xequiv apmod{!n}, xequiv bpmod{! m} $ is solvable for all values of $,a,b.,$ This is true in your example $iff 1 = gcd(n,npm a) = gcd(n,a).$



However, when $,d > 1,$a solution exists iff $,bequiv apmod{!d},,$ and this does place constraints on the values $,b,$ that $,x,$ can take $!bmod m$. Then $,b,$ is determined uniquely iff $,mmid diff mmid n $ (i.e. iff $,npm amid n,$ in your example), hence $,xbmod m = (xbmod n)bmod m,$ is uniquely determined. This arises frequently, e.g. knowing $, u = xbmod 10 = $ units digit of $x$ we can compute its parity by taking the parity of its units digit, i.e. $ xbmod 2 = (xbmod 10)bmod 2 = ubmod 2,,$ e.g. $,123bmod 2 = 3bmod 2 = 1$.






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$endgroup$





















    1












    $begingroup$

    For the specific case $22!$, the answer is easy: since $4 leq 22$ and $6 leq 22$ you have that $22! = 0 pmod{24}$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.



    The answer to your question is no, and here is an illustrative example: suppose you know that $x = 5 pmod{10}$. Then you could have, among many options, that $x = 5$, or that $x = 15$. In the former case, $x = 5 pmod{12}$. In the latter case, $x = 3 pmod{12}$.






    share|cite|improve this answer









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    • $begingroup$
      But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
      $endgroup$
      – Bill Dubuque
      Jan 8 at 20:24











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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

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    active

    oldest

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    1












    $begingroup$

    If $,d := gcd(n,m)=1,$ then there is no relationship between $,a = xbmod n,$ and $, b = xbmod m,$ since CRT $,Rightarrow,xequiv apmod{!n}, xequiv bpmod{! m} $ is solvable for all values of $,a,b.,$ This is true in your example $iff 1 = gcd(n,npm a) = gcd(n,a).$



    However, when $,d > 1,$a solution exists iff $,bequiv apmod{!d},,$ and this does place constraints on the values $,b,$ that $,x,$ can take $!bmod m$. Then $,b,$ is determined uniquely iff $,mmid diff mmid n $ (i.e. iff $,npm amid n,$ in your example), hence $,xbmod m = (xbmod n)bmod m,$ is uniquely determined. This arises frequently, e.g. knowing $, u = xbmod 10 = $ units digit of $x$ we can compute its parity by taking the parity of its units digit, i.e. $ xbmod 2 = (xbmod 10)bmod 2 = ubmod 2,,$ e.g. $,123bmod 2 = 3bmod 2 = 1$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      If $,d := gcd(n,m)=1,$ then there is no relationship between $,a = xbmod n,$ and $, b = xbmod m,$ since CRT $,Rightarrow,xequiv apmod{!n}, xequiv bpmod{! m} $ is solvable for all values of $,a,b.,$ This is true in your example $iff 1 = gcd(n,npm a) = gcd(n,a).$



      However, when $,d > 1,$a solution exists iff $,bequiv apmod{!d},,$ and this does place constraints on the values $,b,$ that $,x,$ can take $!bmod m$. Then $,b,$ is determined uniquely iff $,mmid diff mmid n $ (i.e. iff $,npm amid n,$ in your example), hence $,xbmod m = (xbmod n)bmod m,$ is uniquely determined. This arises frequently, e.g. knowing $, u = xbmod 10 = $ units digit of $x$ we can compute its parity by taking the parity of its units digit, i.e. $ xbmod 2 = (xbmod 10)bmod 2 = ubmod 2,,$ e.g. $,123bmod 2 = 3bmod 2 = 1$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        If $,d := gcd(n,m)=1,$ then there is no relationship between $,a = xbmod n,$ and $, b = xbmod m,$ since CRT $,Rightarrow,xequiv apmod{!n}, xequiv bpmod{! m} $ is solvable for all values of $,a,b.,$ This is true in your example $iff 1 = gcd(n,npm a) = gcd(n,a).$



        However, when $,d > 1,$a solution exists iff $,bequiv apmod{!d},,$ and this does place constraints on the values $,b,$ that $,x,$ can take $!bmod m$. Then $,b,$ is determined uniquely iff $,mmid diff mmid n $ (i.e. iff $,npm amid n,$ in your example), hence $,xbmod m = (xbmod n)bmod m,$ is uniquely determined. This arises frequently, e.g. knowing $, u = xbmod 10 = $ units digit of $x$ we can compute its parity by taking the parity of its units digit, i.e. $ xbmod 2 = (xbmod 10)bmod 2 = ubmod 2,,$ e.g. $,123bmod 2 = 3bmod 2 = 1$.






        share|cite|improve this answer











        $endgroup$



        If $,d := gcd(n,m)=1,$ then there is no relationship between $,a = xbmod n,$ and $, b = xbmod m,$ since CRT $,Rightarrow,xequiv apmod{!n}, xequiv bpmod{! m} $ is solvable for all values of $,a,b.,$ This is true in your example $iff 1 = gcd(n,npm a) = gcd(n,a).$



        However, when $,d > 1,$a solution exists iff $,bequiv apmod{!d},,$ and this does place constraints on the values $,b,$ that $,x,$ can take $!bmod m$. Then $,b,$ is determined uniquely iff $,mmid diff mmid n $ (i.e. iff $,npm amid n,$ in your example), hence $,xbmod m = (xbmod n)bmod m,$ is uniquely determined. This arises frequently, e.g. knowing $, u = xbmod 10 = $ units digit of $x$ we can compute its parity by taking the parity of its units digit, i.e. $ xbmod 2 = (xbmod 10)bmod 2 = ubmod 2,,$ e.g. $,123bmod 2 = 3bmod 2 = 1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 20:33

























        answered Jan 8 at 20:21









        Bill DubuqueBill Dubuque

        209k29191633




        209k29191633























            1












            $begingroup$

            For the specific case $22!$, the answer is easy: since $4 leq 22$ and $6 leq 22$ you have that $22! = 0 pmod{24}$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.



            The answer to your question is no, and here is an illustrative example: suppose you know that $x = 5 pmod{10}$. Then you could have, among many options, that $x = 5$, or that $x = 15$. In the former case, $x = 5 pmod{12}$. In the latter case, $x = 3 pmod{12}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
              $endgroup$
              – Bill Dubuque
              Jan 8 at 20:24
















            1












            $begingroup$

            For the specific case $22!$, the answer is easy: since $4 leq 22$ and $6 leq 22$ you have that $22! = 0 pmod{24}$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.



            The answer to your question is no, and here is an illustrative example: suppose you know that $x = 5 pmod{10}$. Then you could have, among many options, that $x = 5$, or that $x = 15$. In the former case, $x = 5 pmod{12}$. In the latter case, $x = 3 pmod{12}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
              $endgroup$
              – Bill Dubuque
              Jan 8 at 20:24














            1












            1








            1





            $begingroup$

            For the specific case $22!$, the answer is easy: since $4 leq 22$ and $6 leq 22$ you have that $22! = 0 pmod{24}$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.



            The answer to your question is no, and here is an illustrative example: suppose you know that $x = 5 pmod{10}$. Then you could have, among many options, that $x = 5$, or that $x = 15$. In the former case, $x = 5 pmod{12}$. In the latter case, $x = 3 pmod{12}$.






            share|cite|improve this answer









            $endgroup$



            For the specific case $22!$, the answer is easy: since $4 leq 22$ and $6 leq 22$ you have that $22! = 0 pmod{24}$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.



            The answer to your question is no, and here is an illustrative example: suppose you know that $x = 5 pmod{10}$. Then you could have, among many options, that $x = 5$, or that $x = 15$. In the former case, $x = 5 pmod{12}$. In the latter case, $x = 3 pmod{12}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 15:07









            Mees de VriesMees de Vries

            16.4k12654




            16.4k12654












            • $begingroup$
              But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
              $endgroup$
              – Bill Dubuque
              Jan 8 at 20:24


















            • $begingroup$
              But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
              $endgroup$
              – Bill Dubuque
              Jan 8 at 20:24
















            $begingroup$
            But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 20:24




            $begingroup$
            But the answer can be yes too, e.g. $,n=10, n−a=5. $ See my answer for the general case.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 20:24


















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