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











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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)$.










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  • 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
    Nov 19 at 21:17










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    Nov 19 at 21:17






  • 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
    Nov 19 at 21:25















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)$.










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
    Nov 19 at 21:17










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    Nov 19 at 21:17






  • 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
    Nov 19 at 21:25













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)$.










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






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edited Nov 19 at 22:29









Maged Saeed

547315




547315










asked Nov 19 at 21:12









thetha

268115




268115












  • 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
    Nov 19 at 21:17










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    Nov 19 at 21:17






  • 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
    Nov 19 at 21:25


















  • 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
    Nov 19 at 21:17










  • It just means it is divisible by 100. Hint: $101$ is prime
    – Sorfosh
    Nov 19 at 21:17






  • 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
    Nov 19 at 21:25
















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
Nov 19 at 21:17




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
Nov 19 at 21:17












It just means it is divisible by 100. Hint: $101$ is prime
– Sorfosh
Nov 19 at 21:17




It just means it is divisible by 100. Hint: $101$ is prime
– Sorfosh
Nov 19 at 21:17




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
Nov 19 at 21:25




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
Nov 19 at 21:25










3 Answers
3






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1
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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,
$phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$



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






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    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




























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      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





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote



        accepted










        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
          1
          down vote



          accepted










          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
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            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 Nov 19 at 21:35









            Doug M

            43k31753




            43k31753






















                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 Nov 19 at 21:26









                    Maged Saeed

                    547315




                    547315






















                        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 Nov 19 at 21:36









                            Keith Backman

                            731148




                            731148






























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