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







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












$begingroup$


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






share|cite|improve this question















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





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f575971%2fconverting-base-16-digits-to-base-2-and-10-analysis-question%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f575971%2fconverting-base-16-digits-to-base-2-and-10-analysis-question%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more

            Cabo Verde

            Image
              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)
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more