Converting Base 16 digits to base 2 and 10 - Analysis Question












-1












$begingroup$


With base $16$, the digits are denoted as $0, 1,ldots, 9, A,ldots, F$. Let $n = AB3$. Rewrite $n$ with bases $10$, $2$.



I have no clue what this question means and how I should attempt to do this.






number-systems







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$endgroup$












  • $begingroup$
    If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions.
    $endgroup$
    – hardmath
    Nov 21 '13 at 14:03
















-1












$begingroup$


With base $16$, the digits are denoted as $0, 1,ldots, 9, A,ldots, F$. Let $n = AB3$. Rewrite $n$ with bases $10$, $2$.



I have no clue what this question means and how I should attempt to do this.






number-systems







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions.
    $endgroup$
    – hardmath
    Nov 21 '13 at 14:03














-1












-1








-1





$begingroup$


With base $16$, the digits are denoted as $0, 1,ldots, 9, A,ldots, F$. Let $n = AB3$. Rewrite $n$ with bases $10$, $2$.



I have no clue what this question means and how I should attempt to do this.






number-systems







share|cite|improve this question











$endgroup$




With base $16$, the digits are denoted as $0, 1,ldots, 9, A,ldots, F$. Let $n = AB3$. Rewrite $n$ with bases $10$, $2$.



I have no clue what this question means and how I should attempt to do this.






number-systems




number-systems






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 23:35









trw

3331311




3331311










asked Nov 21 '13 at 13:57









Kasun JayasuriyaKasun Jayasuriya

144




144












  • $begingroup$
    If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions.
    $endgroup$
    – hardmath
    Nov 21 '13 at 14:03


















  • $begingroup$
    If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions.
    $endgroup$
    – hardmath
    Nov 21 '13 at 14:03
















$begingroup$
If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions.
$endgroup$
– hardmath
Nov 21 '13 at 14:03




$begingroup$
If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions.
$endgroup$
– hardmath
Nov 21 '13 at 14:03










3 Answers
3






active

oldest

votes


















6












$begingroup$

Hint: a number $n$ in base 10 can be written as follow: $$n=a_i cdot 10^{i} + a_{i-1} cdot 10^{i-1} cdot dots cdot a_1 cdot 10 + a_0 cdot 10^0 $$ where $a_i in {0, 1, dots, 9}$. Similar to this case...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So what's i in this?
    $endgroup$
    – Kasun Jayasuriya
    Nov 21 '13 at 14:10










  • $begingroup$
    It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
    $endgroup$
    – Ross Millikan
    Nov 21 '13 at 14:17



















6












$begingroup$

In base 16, each number is composed of a series of the digits 0-9 and A-F, where, A represents 10, B 11, C 12 and so on up to the maximum for any digit which is F, representing 15.



As such, AB$3$ represents $10$ in the $16^2=256$s column, $11$ in the $16^1=16$s column and 3 in the units column, so in decimal (base 10) that's $10(256)+11(16)+3=2739$.



In general, for a natural number $n$, given a number in base $n$, the digit at the far right tells you how many units, the digit to the left of that how many $n$s, the next digit to the left how many $n^2$s etc. None of the digits in a number in base $n$ is allowed to exceed $n-1$.



For how to convert the number to base 2, see the algorithm presented in Adi Dani's answer.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $$n=(AB3)_{16}=Acdot16^2+Bcdot 16^1+3cdot 16^0=10cdot256+11cdot16+3cdot1=2739$$
    $$2739:2=1369+1$$
    $$1369:2=2cdot684+1$$
    $$684:2=2cdot342+0$$
    $$342:2=2cdot171+0$$
    $$171:2=2cdot85+1$$
    $$85:2=2cdot42+1$$
    $$42:2=2cdot21+0$$
    $$21:2=2cdot10+1$$
    $$10:2=2cdot5+0$$
    $$5:2=2cdot2+1$$
    $$2=2cdot1+0$$



    $$2739=(101010110011)_2$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
      $endgroup$
      – George Tomlinson
      Nov 22 '13 at 20:43








    • 1




      $begingroup$
      I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
      $endgroup$
      – Adi Dani
      Nov 22 '13 at 22:34










    • $begingroup$
      No problems. Oh, yeah: I meant 1369.
      $endgroup$
      – George Tomlinson
      Nov 25 '13 at 17:54












    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Hint: a number $n$ in base 10 can be written as follow: $$n=a_i cdot 10^{i} + a_{i-1} cdot 10^{i-1} cdot dots cdot a_1 cdot 10 + a_0 cdot 10^0 $$ where $a_i in {0, 1, dots, 9}$. Similar to this case...






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So what's i in this?
      $endgroup$
      – Kasun Jayasuriya
      Nov 21 '13 at 14:10










    • $begingroup$
      It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
      $endgroup$
      – Ross Millikan
      Nov 21 '13 at 14:17
















    6












    $begingroup$

    Hint: a number $n$ in base 10 can be written as follow: $$n=a_i cdot 10^{i} + a_{i-1} cdot 10^{i-1} cdot dots cdot a_1 cdot 10 + a_0 cdot 10^0 $$ where $a_i in {0, 1, dots, 9}$. Similar to this case...






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So what's i in this?
      $endgroup$
      – Kasun Jayasuriya
      Nov 21 '13 at 14:10










    • $begingroup$
      It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
      $endgroup$
      – Ross Millikan
      Nov 21 '13 at 14:17














    6












    6








    6





    $begingroup$

    Hint: a number $n$ in base 10 can be written as follow: $$n=a_i cdot 10^{i} + a_{i-1} cdot 10^{i-1} cdot dots cdot a_1 cdot 10 + a_0 cdot 10^0 $$ where $a_i in {0, 1, dots, 9}$. Similar to this case...






    share|cite|improve this answer









    $endgroup$



    Hint: a number $n$ in base 10 can be written as follow: $$n=a_i cdot 10^{i} + a_{i-1} cdot 10^{i-1} cdot dots cdot a_1 cdot 10 + a_0 cdot 10^0 $$ where $a_i in {0, 1, dots, 9}$. Similar to this case...







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 '13 at 14:02









    gangrenegangrene

    915514




    915514












    • $begingroup$
      So what's i in this?
      $endgroup$
      – Kasun Jayasuriya
      Nov 21 '13 at 14:10










    • $begingroup$
      It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
      $endgroup$
      – Ross Millikan
      Nov 21 '13 at 14:17


















    • $begingroup$
      So what's i in this?
      $endgroup$
      – Kasun Jayasuriya
      Nov 21 '13 at 14:10










    • $begingroup$
      It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
      $endgroup$
      – Ross Millikan
      Nov 21 '13 at 14:17
















    $begingroup$
    So what's i in this?
    $endgroup$
    – Kasun Jayasuriya
    Nov 21 '13 at 14:10




    $begingroup$
    So what's i in this?
    $endgroup$
    – Kasun Jayasuriya
    Nov 21 '13 at 14:10












    $begingroup$
    It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
    $endgroup$
    – Ross Millikan
    Nov 21 '13 at 14:17




    $begingroup$
    It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$
    $endgroup$
    – Ross Millikan
    Nov 21 '13 at 14:17











    6












    $begingroup$

    In base 16, each number is composed of a series of the digits 0-9 and A-F, where, A represents 10, B 11, C 12 and so on up to the maximum for any digit which is F, representing 15.



    As such, AB$3$ represents $10$ in the $16^2=256$s column, $11$ in the $16^1=16$s column and 3 in the units column, so in decimal (base 10) that's $10(256)+11(16)+3=2739$.



    In general, for a natural number $n$, given a number in base $n$, the digit at the far right tells you how many units, the digit to the left of that how many $n$s, the next digit to the left how many $n^2$s etc. None of the digits in a number in base $n$ is allowed to exceed $n-1$.



    For how to convert the number to base 2, see the algorithm presented in Adi Dani's answer.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      In base 16, each number is composed of a series of the digits 0-9 and A-F, where, A represents 10, B 11, C 12 and so on up to the maximum for any digit which is F, representing 15.



      As such, AB$3$ represents $10$ in the $16^2=256$s column, $11$ in the $16^1=16$s column and 3 in the units column, so in decimal (base 10) that's $10(256)+11(16)+3=2739$.



      In general, for a natural number $n$, given a number in base $n$, the digit at the far right tells you how many units, the digit to the left of that how many $n$s, the next digit to the left how many $n^2$s etc. None of the digits in a number in base $n$ is allowed to exceed $n-1$.



      For how to convert the number to base 2, see the algorithm presented in Adi Dani's answer.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        In base 16, each number is composed of a series of the digits 0-9 and A-F, where, A represents 10, B 11, C 12 and so on up to the maximum for any digit which is F, representing 15.



        As such, AB$3$ represents $10$ in the $16^2=256$s column, $11$ in the $16^1=16$s column and 3 in the units column, so in decimal (base 10) that's $10(256)+11(16)+3=2739$.



        In general, for a natural number $n$, given a number in base $n$, the digit at the far right tells you how many units, the digit to the left of that how many $n$s, the next digit to the left how many $n^2$s etc. None of the digits in a number in base $n$ is allowed to exceed $n-1$.



        For how to convert the number to base 2, see the algorithm presented in Adi Dani's answer.






        share|cite|improve this answer











        $endgroup$



        In base 16, each number is composed of a series of the digits 0-9 and A-F, where, A represents 10, B 11, C 12 and so on up to the maximum for any digit which is F, representing 15.



        As such, AB$3$ represents $10$ in the $16^2=256$s column, $11$ in the $16^1=16$s column and 3 in the units column, so in decimal (base 10) that's $10(256)+11(16)+3=2739$.



        In general, for a natural number $n$, given a number in base $n$, the digit at the far right tells you how many units, the digit to the left of that how many $n$s, the next digit to the left how many $n^2$s etc. None of the digits in a number in base $n$ is allowed to exceed $n-1$.



        For how to convert the number to base 2, see the algorithm presented in Adi Dani's answer.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 '13 at 19:06

























        answered Nov 21 '13 at 14:11









        George TomlinsonGeorge Tomlinson

        1,216711




        1,216711























            3












            $begingroup$

            $$n=(AB3)_{16}=Acdot16^2+Bcdot 16^1+3cdot 16^0=10cdot256+11cdot16+3cdot1=2739$$
            $$2739:2=1369+1$$
            $$1369:2=2cdot684+1$$
            $$684:2=2cdot342+0$$
            $$342:2=2cdot171+0$$
            $$171:2=2cdot85+1$$
            $$85:2=2cdot42+1$$
            $$42:2=2cdot21+0$$
            $$21:2=2cdot10+1$$
            $$10:2=2cdot5+0$$
            $$5:2=2cdot2+1$$
            $$2=2cdot1+0$$



            $$2739=(101010110011)_2$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
              $endgroup$
              – George Tomlinson
              Nov 22 '13 at 20:43








            • 1




              $begingroup$
              I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
              $endgroup$
              – Adi Dani
              Nov 22 '13 at 22:34










            • $begingroup$
              No problems. Oh, yeah: I meant 1369.
              $endgroup$
              – George Tomlinson
              Nov 25 '13 at 17:54
















            3












            $begingroup$

            $$n=(AB3)_{16}=Acdot16^2+Bcdot 16^1+3cdot 16^0=10cdot256+11cdot16+3cdot1=2739$$
            $$2739:2=1369+1$$
            $$1369:2=2cdot684+1$$
            $$684:2=2cdot342+0$$
            $$342:2=2cdot171+0$$
            $$171:2=2cdot85+1$$
            $$85:2=2cdot42+1$$
            $$42:2=2cdot21+0$$
            $$21:2=2cdot10+1$$
            $$10:2=2cdot5+0$$
            $$5:2=2cdot2+1$$
            $$2=2cdot1+0$$



            $$2739=(101010110011)_2$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
              $endgroup$
              – George Tomlinson
              Nov 22 '13 at 20:43








            • 1




              $begingroup$
              I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
              $endgroup$
              – Adi Dani
              Nov 22 '13 at 22:34










            • $begingroup$
              No problems. Oh, yeah: I meant 1369.
              $endgroup$
              – George Tomlinson
              Nov 25 '13 at 17:54














            3












            3








            3





            $begingroup$

            $$n=(AB3)_{16}=Acdot16^2+Bcdot 16^1+3cdot 16^0=10cdot256+11cdot16+3cdot1=2739$$
            $$2739:2=1369+1$$
            $$1369:2=2cdot684+1$$
            $$684:2=2cdot342+0$$
            $$342:2=2cdot171+0$$
            $$171:2=2cdot85+1$$
            $$85:2=2cdot42+1$$
            $$42:2=2cdot21+0$$
            $$21:2=2cdot10+1$$
            $$10:2=2cdot5+0$$
            $$5:2=2cdot2+1$$
            $$2=2cdot1+0$$



            $$2739=(101010110011)_2$$






            share|cite|improve this answer











            $endgroup$



            $$n=(AB3)_{16}=Acdot16^2+Bcdot 16^1+3cdot 16^0=10cdot256+11cdot16+3cdot1=2739$$
            $$2739:2=1369+1$$
            $$1369:2=2cdot684+1$$
            $$684:2=2cdot342+0$$
            $$342:2=2cdot171+0$$
            $$171:2=2cdot85+1$$
            $$85:2=2cdot42+1$$
            $$42:2=2cdot21+0$$
            $$21:2=2cdot10+1$$
            $$10:2=2cdot5+0$$
            $$5:2=2cdot2+1$$
            $$2=2cdot1+0$$



            $$2739=(101010110011)_2$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 22 '13 at 22:30

























            answered Nov 21 '13 at 14:12









            Adi DaniAdi Dani

            15.3k32246




            15.3k32246












            • $begingroup$
              I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
              $endgroup$
              – George Tomlinson
              Nov 22 '13 at 20:43








            • 1




              $begingroup$
              I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
              $endgroup$
              – Adi Dani
              Nov 22 '13 at 22:34










            • $begingroup$
              No problems. Oh, yeah: I meant 1369.
              $endgroup$
              – George Tomlinson
              Nov 25 '13 at 17:54


















            • $begingroup$
              I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
              $endgroup$
              – George Tomlinson
              Nov 22 '13 at 20:43








            • 1




              $begingroup$
              I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
              $endgroup$
              – Adi Dani
              Nov 22 '13 at 22:34










            • $begingroup$
              No problems. Oh, yeah: I meant 1369.
              $endgroup$
              – George Tomlinson
              Nov 25 '13 at 17:54
















            $begingroup$
            I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
            $endgroup$
            – George Tomlinson
            Nov 22 '13 at 20:43






            $begingroup$
            I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$
            $endgroup$
            – George Tomlinson
            Nov 22 '13 at 20:43






            1




            1




            $begingroup$
            I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
            $endgroup$
            – Adi Dani
            Nov 22 '13 at 22:34




            $begingroup$
            I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34
            $endgroup$
            – Adi Dani
            Nov 22 '13 at 22:34












            $begingroup$
            No problems. Oh, yeah: I meant 1369.
            $endgroup$
            – George Tomlinson
            Nov 25 '13 at 17:54




            $begingroup$
            No problems. Oh, yeah: I meant 1369.
            $endgroup$
            – George Tomlinson
            Nov 25 '13 at 17:54


















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