Power tower last digits












1












$begingroup$


Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



$6^{6^{6^{6^{6^{6}}}}}$



Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






modular-arithmetic problem-solving







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



    $6^{6^{6^{6^{6^{6}}}}}$



    Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






    modular-arithmetic problem-solving







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



      $6^{6^{6^{6^{6^{6}}}}}$



      Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






      modular-arithmetic problem-solving







      share|cite|improve this question









      $endgroup$




      Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



      $6^{6^{6^{6^{6^{6}}}}}$



      Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






      modular-arithmetic problem-solving




      modular-arithmetic problem-solving






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 19:45









      Michael MuntaMichael Munta

      1048




      1048






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47













          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%2f3072436%2fpower-tower-last-digits%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47


















          2












          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47
















          2












          2








          2





          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$



          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 21:01









          Robert IsraelRobert Israel

          323k23212466




          323k23212466












          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47




















          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47


















          $begingroup$
          How did you do this mod $5^n$ without calculator? For example $6^{1156}$
          $endgroup$
          – Michael Munta
          Jan 13 at 21:14






          $begingroup$
          How did you do this mod $5^n$ without calculator? For example $6^{1156}$
          $endgroup$
          – Michael Munta
          Jan 13 at 21:14






          1




          1




          $begingroup$
          I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
          $endgroup$
          – Robert Israel
          Jan 14 at 1:11




          $begingroup$
          I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
          $endgroup$
          – Robert Israel
          Jan 14 at 1:11












          $begingroup$
          And this process of finding multiplicative order, does that have a name? What did you do there really?
          $endgroup$
          – Michael Munta
          Jan 14 at 6:35




          $begingroup$
          And this process of finding multiplicative order, does that have a name? What did you do there really?
          $endgroup$
          – Michael Munta
          Jan 14 at 6:35












          $begingroup$
          Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
          $endgroup$
          – Michael Munta
          Jan 14 at 10:44




          $begingroup$
          Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
          $endgroup$
          – Michael Munta
          Jan 14 at 10:44












          $begingroup$
          $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
          $endgroup$
          – Robert Israel
          Jan 14 at 12:47






          $begingroup$
          $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
          $endgroup$
          – Robert Israel
          Jan 14 at 12:47




















          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%2f3072436%2fpower-tower-last-digits%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

          android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d
          Read more

          SQL update select statement

          Image
          0 I'm trying to update an entry in my database (postgresql). I'm having a problem with updating the query: INSERT INTO prices (price, product_id) SELECT product_price, commn_id FROM products_temp as prod WHERE NOT EXISTS (SELECT * FROM prices od WHERE prod.commn_id = od.product_id) OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) Query works fine when i delete: OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) So it seems to me that it's looping through this operation. My question is, how can I fix it? sql postgresql
          Read more

          'app-layout' is not a known element: how to share Component with different Modules

          Image
          0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule
          Read more