How to find in which number base the operation was done by looking at the corresponding operation in decimal...












5












$begingroup$


$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this



Method 1 : Through equations



assume base be a
$$23_a + 25_a = 51_a $$



$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7



Method 2 : Just logical reasoning



as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .



No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .



To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?



That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .



I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .



Thanks






elementary-number-theory number-systems







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
    $endgroup$
    – egreg
    Jul 26 '13 at 11:29










  • $begingroup$
    I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:04






  • 2




    $begingroup$
    We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
    $endgroup$
    – egreg
    Jul 26 '13 at 20:11
















5












$begingroup$


$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this



Method 1 : Through equations



assume base be a
$$23_a + 25_a = 51_a $$



$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7



Method 2 : Just logical reasoning



as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .



No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .



To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?



That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .



I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .



Thanks






elementary-number-theory number-systems







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
    $endgroup$
    – egreg
    Jul 26 '13 at 11:29










  • $begingroup$
    I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:04






  • 2




    $begingroup$
    We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
    $endgroup$
    – egreg
    Jul 26 '13 at 20:11














5












5








5


2



$begingroup$


$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this



Method 1 : Through equations



assume base be a
$$23_a + 25_a = 51_a $$



$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7



Method 2 : Just logical reasoning



as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .



No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .



To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?



That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .



I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .



Thanks






elementary-number-theory number-systems







share|cite|improve this question









$endgroup$




$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this



Method 1 : Through equations



assume base be a
$$23_a + 25_a = 51_a $$



$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7



Method 2 : Just logical reasoning



as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .



No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .



To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?



That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .



I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .



Thanks






elementary-number-theory number-systems




elementary-number-theory number-systems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 26 '13 at 10:17









Harish KayarohanamHarish Kayarohanam

1,6701924




1,6701924








  • 1




    $begingroup$
    If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
    $endgroup$
    – egreg
    Jul 26 '13 at 11:29










  • $begingroup$
    I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:04






  • 2




    $begingroup$
    We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
    $endgroup$
    – egreg
    Jul 26 '13 at 20:11














  • 1




    $begingroup$
    If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
    $endgroup$
    – egreg
    Jul 26 '13 at 11:29










  • $begingroup$
    I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:04






  • 2




    $begingroup$
    We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
    $endgroup$
    – egreg
    Jul 26 '13 at 20:11








1




1




$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29




$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29












$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04




$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04




2




2




$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11




$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11










2 Answers
2






active

oldest

votes


















1












$begingroup$

In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:07










  • $begingroup$
    @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
    $endgroup$
    – Mark Bennet
    Jul 26 '13 at 20:09










  • $begingroup$
    I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:12



















1












$begingroup$

I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.



For your problem,



begin{align}
[2,3]_a + [2,5]_a &= [5,1]_a \
[4,8]_a &= [5,1]_a \
[4,8]_a &= [4, a+1] \
8 &= a+ 1 \
a &= 7
end{align}






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:07










    • $begingroup$
      @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
      $endgroup$
      – Mark Bennet
      Jul 26 '13 at 20:09










    • $begingroup$
      I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:12
















    1












    $begingroup$

    In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:07










    • $begingroup$
      @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
      $endgroup$
      – Mark Bennet
      Jul 26 '13 at 20:09










    • $begingroup$
      I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:12














    1












    1








    1





    $begingroup$

    In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.






    share|cite|improve this answer









    $endgroup$



    In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 26 '13 at 11:22









    Mark BennetMark Bennet

    81.4k983180




    81.4k983180












    • $begingroup$
      Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:07










    • $begingroup$
      @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
      $endgroup$
      – Mark Bennet
      Jul 26 '13 at 20:09










    • $begingroup$
      I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:12


















    • $begingroup$
      Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:07










    • $begingroup$
      @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
      $endgroup$
      – Mark Bennet
      Jul 26 '13 at 20:09










    • $begingroup$
      I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
      $endgroup$
      – Harish Kayarohanam
      Jul 26 '13 at 20:12
















    $begingroup$
    Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:07




    $begingroup$
    Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:07












    $begingroup$
    @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
    $endgroup$
    – Mark Bennet
    Jul 26 '13 at 20:09




    $begingroup$
    @HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
    $endgroup$
    – Mark Bennet
    Jul 26 '13 at 20:09












    $begingroup$
    I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:12




    $begingroup$
    I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
    $endgroup$
    – Harish Kayarohanam
    Jul 26 '13 at 20:12











    1












    $begingroup$

    I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.



    For your problem,



    begin{align}
    [2,3]_a + [2,5]_a &= [5,1]_a \
    [4,8]_a &= [5,1]_a \
    [4,8]_a &= [4, a+1] \
    8 &= a+ 1 \
    a &= 7
    end{align}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.



      For your problem,



      begin{align}
      [2,3]_a + [2,5]_a &= [5,1]_a \
      [4,8]_a &= [5,1]_a \
      [4,8]_a &= [4, a+1] \
      8 &= a+ 1 \
      a &= 7
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.



        For your problem,



        begin{align}
        [2,3]_a + [2,5]_a &= [5,1]_a \
        [4,8]_a &= [5,1]_a \
        [4,8]_a &= [4, a+1] \
        8 &= a+ 1 \
        a &= 7
        end{align}






        share|cite|improve this answer









        $endgroup$



        I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.



        For your problem,



        begin{align}
        [2,3]_a + [2,5]_a &= [5,1]_a \
        [4,8]_a &= [5,1]_a \
        [4,8]_a &= [4, a+1] \
        8 &= a+ 1 \
        a &= 7
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 8:40









        steven gregorysteven gregory

        18.2k32258




        18.2k32258






























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            Bressuire

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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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            Gyllenstierna

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