Every positive power of 5 appears in the last digits of bigger power of 5












4














Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.



Please give hints towards the right direction and not the full solutions. Thanks!!






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    Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.



    Please give hints towards the right direction and not the full solutions. Thanks!!






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      Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.



      Please give hints towards the right direction and not the full solutions. Thanks!!






      number-theory contest-math







      share|cite|improve this question







      New contributor




      BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.



      Please give hints towards the right direction and not the full solutions. Thanks!!






      number-theory contest-math




      number-theory contest-math






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      asked 1 hour ago









      BrianH

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          2 Answers
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          4














          Fix $n$.
          The cases $n le 3$ can be handled directly.
          We now assume $n > 3$.



          Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.



          You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$



          Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.





          Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.







          share|cite|improve this answer





























            2














            We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits



            So what we need to do is to find $5^Nequiv 5^npmod{10^n}$



            This can be achieved by setting $N=n+phi (2^n)$



            Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$






            share|cite|improve this answer








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            • Could you explain to me how you reached the last two lines?
              – BrianH
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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            Fix $n$.
            The cases $n le 3$ can be handled directly.
            We now assume $n > 3$.



            Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.



            You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$



            Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.





            Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.







            share|cite|improve this answer


























              4














              Fix $n$.
              The cases $n le 3$ can be handled directly.
              We now assume $n > 3$.



              Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.



              You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$



              Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.





              Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.







              share|cite|improve this answer
























                4












                4








                4






                Fix $n$.
                The cases $n le 3$ can be handled directly.
                We now assume $n > 3$.



                Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.



                You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$



                Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.





                Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.







                share|cite|improve this answer












                Fix $n$.
                The cases $n le 3$ can be handled directly.
                We now assume $n > 3$.



                Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.



                You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$



                Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.





                Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 45 mins ago









                angryavian

                38.4k23180




                38.4k23180























                    2














                    We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits



                    So what we need to do is to find $5^Nequiv 5^npmod{10^n}$



                    This can be achieved by setting $N=n+phi (2^n)$



                    Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$






                    share|cite|improve this answer








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                    • Could you explain to me how you reached the last two lines?
                      – BrianH
                      27 mins ago
















                    2














                    We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits



                    So what we need to do is to find $5^Nequiv 5^npmod{10^n}$



                    This can be achieved by setting $N=n+phi (2^n)$



                    Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$






                    share|cite|improve this answer








                    New contributor




                    bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.


















                    • Could you explain to me how you reached the last two lines?
                      – BrianH
                      27 mins ago














                    2












                    2








                    2






                    We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits



                    So what we need to do is to find $5^Nequiv 5^npmod{10^n}$



                    This can be achieved by setting $N=n+phi (2^n)$



                    Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$






                    share|cite|improve this answer








                    New contributor




                    bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits



                    So what we need to do is to find $5^Nequiv 5^npmod{10^n}$



                    This can be achieved by setting $N=n+phi (2^n)$



                    Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$







                    share|cite|improve this answer








                    New contributor




                    bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 40 mins ago









                    bangzheng

                    211




                    211




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                    New contributor





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                    bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    • Could you explain to me how you reached the last two lines?
                      – BrianH
                      27 mins ago


















                    • Could you explain to me how you reached the last two lines?
                      – BrianH
                      27 mins ago
















                    Could you explain to me how you reached the last two lines?
                    – BrianH
                    27 mins ago




                    Could you explain to me how you reached the last two lines?
                    – BrianH
                    27 mins ago










                    BrianH is a new contributor. Be nice, and check out our Code of Conduct.










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