last 2 digits of a sequence

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$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?



Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step




sequences-and-series




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  • The next step is exactly the same as the first step.
    – Arthur
    Aug 2 at 9:33






  • 2




    The last two digits alternate between $07$ and $47$.
    – Mohammad Zuhair Khan
    Aug 2 at 9:35










  • math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
    – lab bhattacharjee
    Aug 2 at 9:36














up vote
4
down vote

favorite
1












$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?



Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step




sequences-and-series




share|cite|improve this question



















  • The next step is exactly the same as the first step.
    – Arthur
    Aug 2 at 9:33






  • 2




    The last two digits alternate between $07$ and $47$.
    – Mohammad Zuhair Khan
    Aug 2 at 9:35










  • math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
    – lab bhattacharjee
    Aug 2 at 9:36












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?



Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step




sequences-and-series




share|cite|improve this question











$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?



Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step





sequences-and-series





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 2 at 9:25









SuperMage1

53718




53718











  • The next step is exactly the same as the first step.
    – Arthur
    Aug 2 at 9:33






  • 2




    The last two digits alternate between $07$ and $47$.
    – Mohammad Zuhair Khan
    Aug 2 at 9:35










  • math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
    – lab bhattacharjee
    Aug 2 at 9:36
















  • The next step is exactly the same as the first step.
    – Arthur
    Aug 2 at 9:33






  • 2




    The last two digits alternate between $07$ and $47$.
    – Mohammad Zuhair Khan
    Aug 2 at 9:35










  • math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
    – lab bhattacharjee
    Aug 2 at 9:36















The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33




The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33




2




2




The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35




The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35












math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36




math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36










2 Answers
2






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oldest

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up vote
4
down vote



accepted










Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.



So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.



Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.



However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.



Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$



So now, it should be obvious which term appears at $n=2013$.






share|cite|improve this answer





















  • Well the algebra certainly beats my trial and error explanation
    – Mohammad Zuhair Khan
    Aug 2 at 9:41

















up vote
0
down vote













The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$



That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$



Starting with 3 and working mod $(100)$ the algorithm goes like,



$$3to 7 to 47 to 07 to 47 to 07 to .....$$



Now you can finish the solution.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.



    So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.



    Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.



    However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.



    Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$



    So now, it should be obvious which term appears at $n=2013$.






    share|cite|improve this answer





















    • Well the algebra certainly beats my trial and error explanation
      – Mohammad Zuhair Khan
      Aug 2 at 9:41














    up vote
    4
    down vote



    accepted










    Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.



    So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.



    Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.



    However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.



    Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$



    So now, it should be obvious which term appears at $n=2013$.






    share|cite|improve this answer





















    • Well the algebra certainly beats my trial and error explanation
      – Mohammad Zuhair Khan
      Aug 2 at 9:41












    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.



    So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.



    Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.



    However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.



    Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$



    So now, it should be obvious which term appears at $n=2013$.






    share|cite|improve this answer













    Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.



    So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.



    Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.



    However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.



    Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$



    So now, it should be obvious which term appears at $n=2013$.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Aug 2 at 9:37









    астон вілла олоф мэллбэрг

    31.5k22463




    31.5k22463











    • Well the algebra certainly beats my trial and error explanation
      – Mohammad Zuhair Khan
      Aug 2 at 9:41
















    • Well the algebra certainly beats my trial and error explanation
      – Mohammad Zuhair Khan
      Aug 2 at 9:41















    Well the algebra certainly beats my trial and error explanation
    – Mohammad Zuhair Khan
    Aug 2 at 9:41




    Well the algebra certainly beats my trial and error explanation
    – Mohammad Zuhair Khan
    Aug 2 at 9:41










    up vote
    0
    down vote













    The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$



    That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$



    Starting with 3 and working mod $(100)$ the algorithm goes like,



    $$3to 7 to 47 to 07 to 47 to 07 to .....$$



    Now you can finish the solution.






    share|cite|improve this answer

























      up vote
      0
      down vote













      The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$



      That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$



      Starting with 3 and working mod $(100)$ the algorithm goes like,



      $$3to 7 to 47 to 07 to 47 to 07 to .....$$



      Now you can finish the solution.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$



        That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$



        Starting with 3 and working mod $(100)$ the algorithm goes like,



        $$3to 7 to 47 to 07 to 47 to 07 to .....$$



        Now you can finish the solution.






        share|cite|improve this answer













        The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$



        That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$



        Starting with 3 and working mod $(100)$ the algorithm goes like,



        $$3to 7 to 47 to 07 to 47 to 07 to .....$$



        Now you can finish the solution.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 9:44









        Mohammad Riazi-Kermani

        27.1k41851




        27.1k41851






















             

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