Numbers $n$ such that $n$ plus the sum of $n$'s digits is $313$












1












$begingroup$


Good morning, everyone. Here is the problem I'm faced with:




The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?




By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?



It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).



I tried as below:



Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to



$$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$



Thus, simplifying, we seek integer solutions to



$$101 x + 11 y + 2 z = 313$$



I am stuck now as to how to solve this.



Thanks to all in advance.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Good morning, everyone. Here is the problem I'm faced with:




    The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?




    By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?



    It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).



    I tried as below:



    Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to



    $$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$



    Thus, simplifying, we seek integer solutions to



    $$101 x + 11 y + 2 z = 313$$



    I am stuck now as to how to solve this.



    Thanks to all in advance.










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Good morning, everyone. Here is the problem I'm faced with:




      The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?




      By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?



      It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).



      I tried as below:



      Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to



      $$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$



      Thus, simplifying, we seek integer solutions to



      $$101 x + 11 y + 2 z = 313$$



      I am stuck now as to how to solve this.



      Thanks to all in advance.










      share|cite|improve this question











      $endgroup$




      Good morning, everyone. Here is the problem I'm faced with:




      The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?




      By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?



      It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).



      I tried as below:



      Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to



      $$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$



      Thus, simplifying, we seek integer solutions to



      $$101 x + 11 y + 2 z = 313$$



      I am stuck now as to how to solve this.



      Thanks to all in advance.







      number-theory elementary-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 7:51









      Eevee Trainer

      7,97021439




      7,97021439










      asked Jan 13 at 7:22









      BlueStarBlueStar

      82




      82






















          4 Answers
          4






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          0












          $begingroup$

          Going from your final conclusion,



          $$101x + 11y + 2z = 313$$



          it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:



          $$x in {1,2,3}$$



          From here, it becomes more processes of elimination.





          Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.





          Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.



          Try $y=9$. This yields $z=6$.



          Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.





          Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.



          As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.





          Thus, we end up with two solutions:



          $$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
          $$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$



          Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to



          $$n + s(n) = 313$$



          where $s(n)$ denotes the sum of the digits of $n$.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$



            Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              You are on the right track with:
              $$overline{xyz}+x+y+z=101x+11y+2z=313.$$
              $xne 1$, because otherwise $11y+2zle 117<212$.



              $x=2$, then:
              $$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$






              share|cite|improve this answer









              $endgroup$





















                1












                $begingroup$

                $$101x+11y+2z=313$$



                where $x,y,z$ are numbers between $0$ and $9$.



                If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $



                If $x=2$, then we have $$11y+2z=111$$



                Since $zle 9$, we have $$11y =111-2z ge 93$$



                Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$



                Hence $296$ is a solution.



                If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.



                The solutions are $305$ and $296$.






                share|cite|improve this answer









                $endgroup$













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                  4 Answers
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                  0












                  $begingroup$

                  Going from your final conclusion,



                  $$101x + 11y + 2z = 313$$



                  it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:



                  $$x in {1,2,3}$$



                  From here, it becomes more processes of elimination.





                  Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.





                  Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.



                  Try $y=9$. This yields $z=6$.



                  Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.





                  Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.



                  As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.





                  Thus, we end up with two solutions:



                  $$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
                  $$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$



                  Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to



                  $$n + s(n) = 313$$



                  where $s(n)$ denotes the sum of the digits of $n$.






                  share|cite|improve this answer











                  $endgroup$


















                    0












                    $begingroup$

                    Going from your final conclusion,



                    $$101x + 11y + 2z = 313$$



                    it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:



                    $$x in {1,2,3}$$



                    From here, it becomes more processes of elimination.





                    Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.





                    Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.



                    Try $y=9$. This yields $z=6$.



                    Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.





                    Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.



                    As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.





                    Thus, we end up with two solutions:



                    $$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
                    $$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$



                    Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to



                    $$n + s(n) = 313$$



                    where $s(n)$ denotes the sum of the digits of $n$.






                    share|cite|improve this answer











                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      Going from your final conclusion,



                      $$101x + 11y + 2z = 313$$



                      it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:



                      $$x in {1,2,3}$$



                      From here, it becomes more processes of elimination.





                      Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.





                      Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.



                      Try $y=9$. This yields $z=6$.



                      Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.





                      Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.



                      As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.





                      Thus, we end up with two solutions:



                      $$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
                      $$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$



                      Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to



                      $$n + s(n) = 313$$



                      where $s(n)$ denotes the sum of the digits of $n$.






                      share|cite|improve this answer











                      $endgroup$



                      Going from your final conclusion,



                      $$101x + 11y + 2z = 313$$



                      it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:



                      $$x in {1,2,3}$$



                      From here, it becomes more processes of elimination.





                      Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.





                      Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.



                      Try $y=9$. This yields $z=6$.



                      Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.





                      Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.



                      As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.





                      Thus, we end up with two solutions:



                      $$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
                      $$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$



                      Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to



                      $$n + s(n) = 313$$



                      where $s(n)$ denotes the sum of the digits of $n$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 13 at 7:52

























                      answered Jan 13 at 7:46









                      Eevee TrainerEevee Trainer

                      7,97021439




                      7,97021439























                          2












                          $begingroup$

                          The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$



                          Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$



                            Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$



                              Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.






                              share|cite|improve this answer









                              $endgroup$



                              The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$



                              Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 13 at 8:26









                              Mark BennetMark Bennet

                              81.6k984181




                              81.6k984181























                                  1












                                  $begingroup$

                                  You are on the right track with:
                                  $$overline{xyz}+x+y+z=101x+11y+2z=313.$$
                                  $xne 1$, because otherwise $11y+2zle 117<212$.



                                  $x=2$, then:
                                  $$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    You are on the right track with:
                                    $$overline{xyz}+x+y+z=101x+11y+2z=313.$$
                                    $xne 1$, because otherwise $11y+2zle 117<212$.



                                    $x=2$, then:
                                    $$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      You are on the right track with:
                                      $$overline{xyz}+x+y+z=101x+11y+2z=313.$$
                                      $xne 1$, because otherwise $11y+2zle 117<212$.



                                      $x=2$, then:
                                      $$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      You are on the right track with:
                                      $$overline{xyz}+x+y+z=101x+11y+2z=313.$$
                                      $xne 1$, because otherwise $11y+2zle 117<212$.



                                      $x=2$, then:
                                      $$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 13 at 7:35









                                      farruhotafarruhota

                                      21.1k2841




                                      21.1k2841























                                          1












                                          $begingroup$

                                          $$101x+11y+2z=313$$



                                          where $x,y,z$ are numbers between $0$ and $9$.



                                          If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $



                                          If $x=2$, then we have $$11y+2z=111$$



                                          Since $zle 9$, we have $$11y =111-2z ge 93$$



                                          Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$



                                          Hence $296$ is a solution.



                                          If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.



                                          The solutions are $305$ and $296$.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            1












                                            $begingroup$

                                            $$101x+11y+2z=313$$



                                            where $x,y,z$ are numbers between $0$ and $9$.



                                            If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $



                                            If $x=2$, then we have $$11y+2z=111$$



                                            Since $zle 9$, we have $$11y =111-2z ge 93$$



                                            Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$



                                            Hence $296$ is a solution.



                                            If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.



                                            The solutions are $305$ and $296$.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              1












                                              1








                                              1





                                              $begingroup$

                                              $$101x+11y+2z=313$$



                                              where $x,y,z$ are numbers between $0$ and $9$.



                                              If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $



                                              If $x=2$, then we have $$11y+2z=111$$



                                              Since $zle 9$, we have $$11y =111-2z ge 93$$



                                              Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$



                                              Hence $296$ is a solution.



                                              If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.



                                              The solutions are $305$ and $296$.






                                              share|cite|improve this answer









                                              $endgroup$



                                              $$101x+11y+2z=313$$



                                              where $x,y,z$ are numbers between $0$ and $9$.



                                              If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $



                                              If $x=2$, then we have $$11y+2z=111$$



                                              Since $zle 9$, we have $$11y =111-2z ge 93$$



                                              Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$



                                              Hence $296$ is a solution.



                                              If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.



                                              The solutions are $305$ and $296$.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Jan 13 at 7:36









                                              Siong Thye GohSiong Thye Goh

                                              103k1468119




                                              103k1468119






























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