Find a number such that $S(n)+P(n)=n$












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Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits



I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?






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    Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits



    I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?






    elementary-number-theory







    share|cite|improve this question



























      0












      0








      0







      Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits



      I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?






      elementary-number-theory







      share|cite|improve this question















      Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits



      I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?






      elementary-number-theory




      elementary-number-theory






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      edited Nov 24 at 9:49









      greedoid

      37.7k114794




      37.7k114794










      asked Nov 24 at 5:13









      Marko Škorić

      70310




      70310






















          3 Answers
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          1














          A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:



          $$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$



          and



          $$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$



          In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction



          $$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$



          (Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)






          share|cite|improve this answer





























            1














            Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then



            $$S(n)leq 10(k-1)+d$$



            and



            $$P(n)leq d9^{k-1},$$



            so



            $$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$



            $$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$



            $$10^{k-1}-9^{k-1}-1leq 10(k-1).$$



            Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?






            share|cite|improve this answer





























              1














              Hint:



              Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$



              Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...






              share|cite|improve this answer























                Your Answer





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                3 Answers
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                3 Answers
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                active

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                active

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                1














                A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:



                $$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$



                and



                $$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$



                In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction



                $$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$



                (Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)






                share|cite|improve this answer


























                  1














                  A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:



                  $$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$



                  and



                  $$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$



                  In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction



                  $$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$



                  (Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)






                  share|cite|improve this answer
























                    1












                    1








                    1






                    A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:



                    $$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$



                    and



                    $$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$



                    In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction



                    $$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$



                    (Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)






                    share|cite|improve this answer












                    A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:



                    $$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$



                    and



                    $$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$



                    In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction



                    $$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$



                    (Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 at 12:04









                    Barry Cipra

                    59k653123




                    59k653123























                        1














                        Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then



                        $$S(n)leq 10(k-1)+d$$



                        and



                        $$P(n)leq d9^{k-1},$$



                        so



                        $$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$



                        $$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$



                        $$10^{k-1}-9^{k-1}-1leq 10(k-1).$$



                        Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?






                        share|cite|improve this answer


























                          1














                          Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then



                          $$S(n)leq 10(k-1)+d$$



                          and



                          $$P(n)leq d9^{k-1},$$



                          so



                          $$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$



                          $$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$



                          $$10^{k-1}-9^{k-1}-1leq 10(k-1).$$



                          Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then



                            $$S(n)leq 10(k-1)+d$$



                            and



                            $$P(n)leq d9^{k-1},$$



                            so



                            $$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$



                            $$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$



                            $$10^{k-1}-9^{k-1}-1leq 10(k-1).$$



                            Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?






                            share|cite|improve this answer












                            Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then



                            $$S(n)leq 10(k-1)+d$$



                            and



                            $$P(n)leq d9^{k-1},$$



                            so



                            $$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$



                            $$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$



                            $$10^{k-1}-9^{k-1}-1leq 10(k-1).$$



                            Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 at 5:22









                            Carl Schildkraut

                            11.1k11441




                            11.1k11441























                                1














                                Hint:



                                Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$



                                Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...






                                share|cite|improve this answer




























                                  1














                                  Hint:



                                  Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$



                                  Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...






                                  share|cite|improve this answer


























                                    1












                                    1








                                    1






                                    Hint:



                                    Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$



                                    Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...






                                    share|cite|improve this answer














                                    Hint:



                                    Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$



                                    Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 24 at 10:47

























                                    answered Nov 24 at 9:48









                                    greedoid

                                    37.7k114794




                                    37.7k114794






























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