To find smallest $n$ such that $2^{n} equiv 111 pmod{125} $











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How to find the smallest natural number $n$ such that $2^{n} equiv 111 pmod{125} $.




If we consider $pmod{5}$ then $2^n equiv 1 pmod{5}$. For $n=4k+l$ where $l in left{ 0, 1, 2, 3right}$ we get $16^k 2^l equiv 1 pmod{5}$ and $2^l equiv 1 pmod{5}$, which leads to $l=0$. Therefore, equation reduces to $16^k equiv 111 pmod{125}$.



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elementary-number-theory







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  • 1




    Sage says $n=36$
    – Oldboy
    2 hours ago










  • I think $2^nequiv 111equiv -14pmod{125}$ implies that $2^nequiv 11pmod{25}$
    – 1ENİGMA1
    1 hour ago










  • Maybe, similar implicity used here:math.stackexchange.com/questions/1133616/…
    – 1ENİGMA1
    59 mins ago










  • @1ENİGMA1 That questions solves $n^3equiv888pmod{1000}$, which is quite a different question.
    – Servaes
    8 mins ago















up vote
1
down vote

favorite
1













How to find the smallest natural number $n$ such that $2^{n} equiv 111 pmod{125} $.




If we consider $pmod{5}$ then $2^n equiv 1 pmod{5}$. For $n=4k+l$ where $l in left{ 0, 1, 2, 3right}$ we get $16^k 2^l equiv 1 pmod{5}$ and $2^l equiv 1 pmod{5}$, which leads to $l=0$. Therefore, equation reduces to $16^k equiv 111 pmod{125}$.



What to do next?






elementary-number-theory







share|cite|improve this question


















  • 1




    Sage says $n=36$
    – Oldboy
    2 hours ago










  • I think $2^nequiv 111equiv -14pmod{125}$ implies that $2^nequiv 11pmod{25}$
    – 1ENİGMA1
    1 hour ago










  • Maybe, similar implicity used here:math.stackexchange.com/questions/1133616/…
    – 1ENİGMA1
    59 mins ago










  • @1ENİGMA1 That questions solves $n^3equiv888pmod{1000}$, which is quite a different question.
    – Servaes
    8 mins ago













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






How to find the smallest natural number $n$ such that $2^{n} equiv 111 pmod{125} $.




If we consider $pmod{5}$ then $2^n equiv 1 pmod{5}$. For $n=4k+l$ where $l in left{ 0, 1, 2, 3right}$ we get $16^k 2^l equiv 1 pmod{5}$ and $2^l equiv 1 pmod{5}$, which leads to $l=0$. Therefore, equation reduces to $16^k equiv 111 pmod{125}$.



What to do next?






elementary-number-theory







share|cite|improve this question














How to find the smallest natural number $n$ such that $2^{n} equiv 111 pmod{125} $.




If we consider $pmod{5}$ then $2^n equiv 1 pmod{5}$. For $n=4k+l$ where $l in left{ 0, 1, 2, 3right}$ we get $16^k 2^l equiv 1 pmod{5}$ and $2^l equiv 1 pmod{5}$, which leads to $l=0$. Therefore, equation reduces to $16^k equiv 111 pmod{125}$.



What to do next?






elementary-number-theory




elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 2 hours ago









Ghartal

2,85511335




2,85511335








  • 1




    Sage says $n=36$
    – Oldboy
    2 hours ago










  • I think $2^nequiv 111equiv -14pmod{125}$ implies that $2^nequiv 11pmod{25}$
    – 1ENİGMA1
    1 hour ago










  • Maybe, similar implicity used here:math.stackexchange.com/questions/1133616/…
    – 1ENİGMA1
    59 mins ago










  • @1ENİGMA1 That questions solves $n^3equiv888pmod{1000}$, which is quite a different question.
    – Servaes
    8 mins ago














  • 1




    Sage says $n=36$
    – Oldboy
    2 hours ago










  • I think $2^nequiv 111equiv -14pmod{125}$ implies that $2^nequiv 11pmod{25}$
    – 1ENİGMA1
    1 hour ago










  • Maybe, similar implicity used here:math.stackexchange.com/questions/1133616/…
    – 1ENİGMA1
    59 mins ago










  • @1ENİGMA1 That questions solves $n^3equiv888pmod{1000}$, which is quite a different question.
    – Servaes
    8 mins ago








1




1




Sage says $n=36$
– Oldboy
2 hours ago




Sage says $n=36$
– Oldboy
2 hours ago












I think $2^nequiv 111equiv -14pmod{125}$ implies that $2^nequiv 11pmod{25}$
– 1ENİGMA1
1 hour ago




I think $2^nequiv 111equiv -14pmod{125}$ implies that $2^nequiv 11pmod{25}$
– 1ENİGMA1
1 hour ago












Maybe, similar implicity used here:math.stackexchange.com/questions/1133616/…
– 1ENİGMA1
59 mins ago




Maybe, similar implicity used here:math.stackexchange.com/questions/1133616/…
– 1ENİGMA1
59 mins ago












@1ENİGMA1 That questions solves $n^3equiv888pmod{1000}$, which is quite a different question.
– Servaes
8 mins ago




@1ENİGMA1 That questions solves $n^3equiv888pmod{1000}$, which is quite a different question.
– Servaes
8 mins ago










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Next solve $16^kequiv11pmod{25}$, which yields $kequiv4pmod{5}$. Set $k=5m+4$ so that
$$16^kequiv16^4times(16^5)^mequiv36times76^mpmod{125}.$$
Because $36times66equiv1pmod{125}$ we now want to solve
$$76^mequiv66times111equiv76pmod{125},$$
which shows that $m=1$ will do, corresponding to $n=36$.






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    Next solve $16^kequiv11pmod{25}$, which yields $kequiv4pmod{5}$. Set $k=5m+4$ so that
    $$16^kequiv16^4times(16^5)^mequiv36times76^mpmod{125}.$$
    Because $36times66equiv1pmod{125}$ we now want to solve
    $$76^mequiv66times111equiv76pmod{125},$$
    which shows that $m=1$ will do, corresponding to $n=36$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Next solve $16^kequiv11pmod{25}$, which yields $kequiv4pmod{5}$. Set $k=5m+4$ so that
      $$16^kequiv16^4times(16^5)^mequiv36times76^mpmod{125}.$$
      Because $36times66equiv1pmod{125}$ we now want to solve
      $$76^mequiv66times111equiv76pmod{125},$$
      which shows that $m=1$ will do, corresponding to $n=36$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Next solve $16^kequiv11pmod{25}$, which yields $kequiv4pmod{5}$. Set $k=5m+4$ so that
        $$16^kequiv16^4times(16^5)^mequiv36times76^mpmod{125}.$$
        Because $36times66equiv1pmod{125}$ we now want to solve
        $$76^mequiv66times111equiv76pmod{125},$$
        which shows that $m=1$ will do, corresponding to $n=36$.






        share|cite|improve this answer












        Next solve $16^kequiv11pmod{25}$, which yields $kequiv4pmod{5}$. Set $k=5m+4$ so that
        $$16^kequiv16^4times(16^5)^mequiv36times76^mpmod{125}.$$
        Because $36times66equiv1pmod{125}$ we now want to solve
        $$76^mequiv66times111equiv76pmod{125},$$
        which shows that $m=1$ will do, corresponding to $n=36$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 31 mins ago









        Servaes

        20.5k33789




        20.5k33789






























             

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