There a infinity numbers of $n$ such that $ phi (n) equiv 0 pmod{100} $











up vote
0
down vote

favorite












I have no clue what this part: $ phi (n) equiv 0 pmod {100} $ means.
$0 pmod {100}$ means I have an equivalence class $[0]$ in $mathbb{Z}$. This also means I have $100, 200, 300 ,cdots$ as the value of $ phi (n)$.






elementary-number-theory totient-function







share|cite|improve this question
























  • Well, it just means that you are to show that there are infinitely many natural numbers $n$ such that $100,|, varphi(n)$. For a hint note that $100=2^2times 5^2$ so, using the standard formulas for $varphi(n)$ try to build large families of good $n$.
    – lulu
    2 days ago










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    2 days ago






  • 1




    I think the toughest part of this question is to show that there exists at least one number $n$ such that $100| phi(n)$
    – Doug M
    2 days ago















up vote
0
down vote

favorite












I have no clue what this part: $ phi (n) equiv 0 pmod {100} $ means.
$0 pmod {100}$ means I have an equivalence class $[0]$ in $mathbb{Z}$. This also means I have $100, 200, 300 ,cdots$ as the value of $ phi (n)$.






elementary-number-theory totient-function







share|cite|improve this question
























  • Well, it just means that you are to show that there are infinitely many natural numbers $n$ such that $100,|, varphi(n)$. For a hint note that $100=2^2times 5^2$ so, using the standard formulas for $varphi(n)$ try to build large families of good $n$.
    – lulu
    2 days ago










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    2 days ago






  • 1




    I think the toughest part of this question is to show that there exists at least one number $n$ such that $100| phi(n)$
    – Doug M
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have no clue what this part: $ phi (n) equiv 0 pmod {100} $ means.
$0 pmod {100}$ means I have an equivalence class $[0]$ in $mathbb{Z}$. This also means I have $100, 200, 300 ,cdots$ as the value of $ phi (n)$.






elementary-number-theory totient-function







share|cite|improve this question















I have no clue what this part: $ phi (n) equiv 0 pmod {100} $ means.
$0 pmod {100}$ means I have an equivalence class $[0]$ in $mathbb{Z}$. This also means I have $100, 200, 300 ,cdots$ as the value of $ phi (n)$.






elementary-number-theory totient-function




elementary-number-theory totient-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Maged Saeed

428315




428315










asked 2 days ago









thetha

266115




266115












  • Well, it just means that you are to show that there are infinitely many natural numbers $n$ such that $100,|, varphi(n)$. For a hint note that $100=2^2times 5^2$ so, using the standard formulas for $varphi(n)$ try to build large families of good $n$.
    – lulu
    2 days ago










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    2 days ago






  • 1




    I think the toughest part of this question is to show that there exists at least one number $n$ such that $100| phi(n)$
    – Doug M
    2 days ago


















  • Well, it just means that you are to show that there are infinitely many natural numbers $n$ such that $100,|, varphi(n)$. For a hint note that $100=2^2times 5^2$ so, using the standard formulas for $varphi(n)$ try to build large families of good $n$.
    – lulu
    2 days ago










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    2 days ago






  • 1




    I think the toughest part of this question is to show that there exists at least one number $n$ such that $100| phi(n)$
    – Doug M
    2 days ago
















Well, it just means that you are to show that there are infinitely many natural numbers $n$ such that $100,|, varphi(n)$. For a hint note that $100=2^2times 5^2$ so, using the standard formulas for $varphi(n)$ try to build large families of good $n$.
– lulu
2 days ago




Well, it just means that you are to show that there are infinitely many natural numbers $n$ such that $100,|, varphi(n)$. For a hint note that $100=2^2times 5^2$ so, using the standard formulas for $varphi(n)$ try to build large families of good $n$.
– lulu
2 days ago












It just means it is divisible by 100. Hint: $101$ is prime
– Sorfosh
2 days ago




It just means it is divisible by 100. Hint: $101$ is prime
– Sorfosh
2 days ago




1




1




I think the toughest part of this question is to show that there exists at least one number $n$ such that $100| phi(n)$
– Doug M
2 days ago




I think the toughest part of this question is to show that there exists at least one number $n$ such that $100| phi(n)$
– Doug M
2 days ago










3 Answers
3






active

oldest

votes

















up vote
0
down vote













You may take this as a Hint:



Consider this sequence, does it have this property?



$1000,10000,100000,1000000,cdots$. Or equivalently, $10^k, k>2, kin mathbb{Z}.$ In fact, values of $phi(10^k)$ are multiples of $400$,



can you show Why?






share|cite|improve this answer




























    up vote
    0
    down vote













    All natural numbers n have a decomposition into prime factors. (Fundamental theorem of Arithmetic)



    $n = p_1^rp_2^scdots p_{i}^z$



    $phi(n) = n (frac {p_1 -1}{p_1})(frac {p_2 -1}{p_2})cdots(frac {p_i -1}{p_i})$



    If $100| phi(n)$



    Then $5^3$ must be a factor of $n$



    In fact $phi(5^3) = 4cdot 5^2 = 100$



    There exists at least one $n$ such that $100|phi(n)$



    For higher powers of $5,$ e.g. $phi(5^4) = 500$



    And for factor the factor 2,
    $phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$



    $forall jge 0, kge 3, phi(2^j5^k) equiv 0pmod {100}$






    share|cite|improve this answer




























      up vote
      0
      down vote













      Since $phi(ab)=phi(a)phi(b)$ for coprime $a,b$, it is sufficient to find some $a$ such that $phi(a)=100$ and then all numbers $n=ab$ give $phi(ab)equiv 0mod{100}$. Try $a=5^3$.






      share|cite|improve this answer





















        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',
        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%2f3005531%2fthere-a-infinity-numbers-of-n-such-that-phi-n-equiv-0-pmod100%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








        up vote
        0
        down vote













        You may take this as a Hint:



        Consider this sequence, does it have this property?



        $1000,10000,100000,1000000,cdots$. Or equivalently, $10^k, k>2, kin mathbb{Z}.$ In fact, values of $phi(10^k)$ are multiples of $400$,



        can you show Why?






        share|cite|improve this answer

























          up vote
          0
          down vote













          You may take this as a Hint:



          Consider this sequence, does it have this property?



          $1000,10000,100000,1000000,cdots$. Or equivalently, $10^k, k>2, kin mathbb{Z}.$ In fact, values of $phi(10^k)$ are multiples of $400$,



          can you show Why?






          share|cite|improve this answer























            up vote
            0
            down vote










            up vote
            0
            down vote









            You may take this as a Hint:



            Consider this sequence, does it have this property?



            $1000,10000,100000,1000000,cdots$. Or equivalently, $10^k, k>2, kin mathbb{Z}.$ In fact, values of $phi(10^k)$ are multiples of $400$,



            can you show Why?






            share|cite|improve this answer












            You may take this as a Hint:



            Consider this sequence, does it have this property?



            $1000,10000,100000,1000000,cdots$. Or equivalently, $10^k, k>2, kin mathbb{Z}.$ In fact, values of $phi(10^k)$ are multiples of $400$,



            can you show Why?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Maged Saeed

            428315




            428315






















                up vote
                0
                down vote













                All natural numbers n have a decomposition into prime factors. (Fundamental theorem of Arithmetic)



                $n = p_1^rp_2^scdots p_{i}^z$



                $phi(n) = n (frac {p_1 -1}{p_1})(frac {p_2 -1}{p_2})cdots(frac {p_i -1}{p_i})$



                If $100| phi(n)$



                Then $5^3$ must be a factor of $n$



                In fact $phi(5^3) = 4cdot 5^2 = 100$



                There exists at least one $n$ such that $100|phi(n)$



                For higher powers of $5,$ e.g. $phi(5^4) = 500$



                And for factor the factor 2,
                $phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$



                $forall jge 0, kge 3, phi(2^j5^k) equiv 0pmod {100}$






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  All natural numbers n have a decomposition into prime factors. (Fundamental theorem of Arithmetic)



                  $n = p_1^rp_2^scdots p_{i}^z$



                  $phi(n) = n (frac {p_1 -1}{p_1})(frac {p_2 -1}{p_2})cdots(frac {p_i -1}{p_i})$



                  If $100| phi(n)$



                  Then $5^3$ must be a factor of $n$



                  In fact $phi(5^3) = 4cdot 5^2 = 100$



                  There exists at least one $n$ such that $100|phi(n)$



                  For higher powers of $5,$ e.g. $phi(5^4) = 500$



                  And for factor the factor 2,
                  $phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$



                  $forall jge 0, kge 3, phi(2^j5^k) equiv 0pmod {100}$






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    All natural numbers n have a decomposition into prime factors. (Fundamental theorem of Arithmetic)



                    $n = p_1^rp_2^scdots p_{i}^z$



                    $phi(n) = n (frac {p_1 -1}{p_1})(frac {p_2 -1}{p_2})cdots(frac {p_i -1}{p_i})$



                    If $100| phi(n)$



                    Then $5^3$ must be a factor of $n$



                    In fact $phi(5^3) = 4cdot 5^2 = 100$



                    There exists at least one $n$ such that $100|phi(n)$



                    For higher powers of $5,$ e.g. $phi(5^4) = 500$



                    And for factor the factor 2,
                    $phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$



                    $forall jge 0, kge 3, phi(2^j5^k) equiv 0pmod {100}$






                    share|cite|improve this answer












                    All natural numbers n have a decomposition into prime factors. (Fundamental theorem of Arithmetic)



                    $n = p_1^rp_2^scdots p_{i}^z$



                    $phi(n) = n (frac {p_1 -1}{p_1})(frac {p_2 -1}{p_2})cdots(frac {p_i -1}{p_i})$



                    If $100| phi(n)$



                    Then $5^3$ must be a factor of $n$



                    In fact $phi(5^3) = 4cdot 5^2 = 100$



                    There exists at least one $n$ such that $100|phi(n)$



                    For higher powers of $5,$ e.g. $phi(5^4) = 500$



                    And for factor the factor 2,
                    $phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$



                    $forall jge 0, kge 3, phi(2^j5^k) equiv 0pmod {100}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    Doug M

                    42.6k31752




                    42.6k31752






















                        up vote
                        0
                        down vote













                        Since $phi(ab)=phi(a)phi(b)$ for coprime $a,b$, it is sufficient to find some $a$ such that $phi(a)=100$ and then all numbers $n=ab$ give $phi(ab)equiv 0mod{100}$. Try $a=5^3$.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Since $phi(ab)=phi(a)phi(b)$ for coprime $a,b$, it is sufficient to find some $a$ such that $phi(a)=100$ and then all numbers $n=ab$ give $phi(ab)equiv 0mod{100}$. Try $a=5^3$.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Since $phi(ab)=phi(a)phi(b)$ for coprime $a,b$, it is sufficient to find some $a$ such that $phi(a)=100$ and then all numbers $n=ab$ give $phi(ab)equiv 0mod{100}$. Try $a=5^3$.






                            share|cite|improve this answer












                            Since $phi(ab)=phi(a)phi(b)$ for coprime $a,b$, it is sufficient to find some $a$ such that $phi(a)=100$ and then all numbers $n=ab$ give $phi(ab)equiv 0mod{100}$. Try $a=5^3$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            Keith Backman

                            686148




                            686148






























                                 

                                draft saved


                                draft discarded



















































                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005531%2fthere-a-infinity-numbers-of-n-such-that-phi-n-equiv-0-pmod100%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

                                MongoDB - Not Authorized To Execute Command

                                Image
                                .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                                Read more

                                Npm cannot find a required file even through it is in the searched directory

                                Image
                                up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
                                Read more

                                in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                                Image
                                1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                                Read more