Factors in a different base $ 2b^2!+!9b!+!7,mid, 7b^2!+!9b!+!2$












6















Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?




Solution:
enter image description here



But base cannot be negative. Could someone please explain where I am going wrong?






elementary-number-theory divisibility number-systems







share|cite|improve this question





























    6















    Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?




    Solution:
    enter image description here



    But base cannot be negative. Could someone please explain where I am going wrong?






    elementary-number-theory divisibility number-systems







    share|cite|improve this question



























      6












      6








      6


      2






      Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?




      Solution:
      enter image description here



      But base cannot be negative. Could someone please explain where I am going wrong?






      elementary-number-theory divisibility number-systems







      share|cite|improve this question
















      Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?




      Solution:
      enter image description here



      But base cannot be negative. Could someone please explain where I am going wrong?






      elementary-number-theory divisibility number-systems




      elementary-number-theory divisibility number-systems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 9 hours ago









      greedoid

      37.9k114794




      37.9k114794










      asked 10 hours ago









      Aamir Khan

      455




      455






















          4 Answers
          4






          active

          oldest

          votes


















          10














          The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.



          If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.



          If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.



          No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.






          share|cite|improve this answer





















          • Thank you so much. This was really very helpful. :)
            – Aamir Khan
            10 hours ago










          • My pleasure, glad to help.
            – vadim123
            10 hours ago










          • @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
            – Bill Dubuque
            10 hours ago








          • 1




            @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
            – vadim123
            9 hours ago












          • @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
            – Bill Dubuque
            9 hours ago





















          6














          Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it



          $$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$



          Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$






          share|cite|improve this answer































            1














            Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$



            we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$



            so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$



            which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.






            share|cite|improve this answer























            • This answer is of no use?
              – greedoid
              7 hours ago



















            0














            $$2B^2+9B+7mid 7B^2+9B+2$$



            Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.



            Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$



            Writing this out "subtraction-style", we get



            $left.begin{array}{c}
            7 & 9 & 2 \
            2 & 9 & 7 \
            hline
            phantom{4}
            end{array}
            right.
            implies
            left.begin{array}{c}
            6 & (B+8) & (B+2) \
            2 & 9 & 7 \
            hline
            4 & (B-1) & (B-5)
            end{array}
            right.
            $



            So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.



            We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.






            share|cite|improve this answer























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              4 Answers
              4






              active

              oldest

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              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              10














              The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.



              If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.



              If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.



              No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.






              share|cite|improve this answer





















              • Thank you so much. This was really very helpful. :)
                – Aamir Khan
                10 hours ago










              • My pleasure, glad to help.
                – vadim123
                10 hours ago










              • @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
                – Bill Dubuque
                10 hours ago








              • 1




                @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
                – vadim123
                9 hours ago












              • @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
                – Bill Dubuque
                9 hours ago


















              10














              The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.



              If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.



              If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.



              No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.






              share|cite|improve this answer





















              • Thank you so much. This was really very helpful. :)
                – Aamir Khan
                10 hours ago










              • My pleasure, glad to help.
                – vadim123
                10 hours ago










              • @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
                – Bill Dubuque
                10 hours ago








              • 1




                @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
                – vadim123
                9 hours ago












              • @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
                – Bill Dubuque
                9 hours ago
















              10












              10








              10






              The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.



              If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.



              If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.



              No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.






              share|cite|improve this answer












              The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.



              If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.



              If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.



              No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 10 hours ago









              vadim123

              75.5k897189




              75.5k897189












              • Thank you so much. This was really very helpful. :)
                – Aamir Khan
                10 hours ago










              • My pleasure, glad to help.
                – vadim123
                10 hours ago










              • @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
                – Bill Dubuque
                10 hours ago








              • 1




                @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
                – vadim123
                9 hours ago












              • @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
                – Bill Dubuque
                9 hours ago




















              • Thank you so much. This was really very helpful. :)
                – Aamir Khan
                10 hours ago










              • My pleasure, glad to help.
                – vadim123
                10 hours ago










              • @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
                – Bill Dubuque
                10 hours ago








              • 1




                @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
                – vadim123
                9 hours ago












              • @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
                – Bill Dubuque
                9 hours ago


















              Thank you so much. This was really very helpful. :)
              – Aamir Khan
              10 hours ago




              Thank you so much. This was really very helpful. :)
              – Aamir Khan
              10 hours ago












              My pleasure, glad to help.
              – vadim123
              10 hours ago




              My pleasure, glad to help.
              – vadim123
              10 hours ago












              @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
              – Bill Dubuque
              10 hours ago






              @AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
              – Bill Dubuque
              10 hours ago






              1




              1




              @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
              – vadim123
              9 hours ago






              @BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
              – vadim123
              9 hours ago














              @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
              – Bill Dubuque
              9 hours ago






              @vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
              – Bill Dubuque
              9 hours ago













              6














              Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it



              $$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$



              Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$






              share|cite|improve this answer




























                6














                Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it



                $$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$



                Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$






                share|cite|improve this answer


























                  6












                  6








                  6






                  Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it



                  $$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$



                  Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$






                  share|cite|improve this answer














                  Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it



                  $$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$



                  Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 10 hours ago









                  Bill Dubuque

                  208k29190628




                  208k29190628























                      1














                      Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$



                      we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$



                      so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$



                      which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.






                      share|cite|improve this answer























                      • This answer is of no use?
                        – greedoid
                        7 hours ago
















                      1














                      Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$



                      we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$



                      so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$



                      which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.






                      share|cite|improve this answer























                      • This answer is of no use?
                        – greedoid
                        7 hours ago














                      1












                      1








                      1






                      Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$



                      we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$



                      so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$



                      which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.






                      share|cite|improve this answer














                      Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$



                      we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$



                      so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$



                      which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 9 hours ago

























                      answered 9 hours ago









                      greedoid

                      37.9k114794




                      37.9k114794












                      • This answer is of no use?
                        – greedoid
                        7 hours ago


















                      • This answer is of no use?
                        – greedoid
                        7 hours ago
















                      This answer is of no use?
                      – greedoid
                      7 hours ago




                      This answer is of no use?
                      – greedoid
                      7 hours ago











                      0














                      $$2B^2+9B+7mid 7B^2+9B+2$$



                      Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.



                      Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$



                      Writing this out "subtraction-style", we get



                      $left.begin{array}{c}
                      7 & 9 & 2 \
                      2 & 9 & 7 \
                      hline
                      phantom{4}
                      end{array}
                      right.
                      implies
                      left.begin{array}{c}
                      6 & (B+8) & (B+2) \
                      2 & 9 & 7 \
                      hline
                      4 & (B-1) & (B-5)
                      end{array}
                      right.
                      $



                      So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.



                      We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.






                      share|cite|improve this answer




























                        0














                        $$2B^2+9B+7mid 7B^2+9B+2$$



                        Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.



                        Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$



                        Writing this out "subtraction-style", we get



                        $left.begin{array}{c}
                        7 & 9 & 2 \
                        2 & 9 & 7 \
                        hline
                        phantom{4}
                        end{array}
                        right.
                        implies
                        left.begin{array}{c}
                        6 & (B+8) & (B+2) \
                        2 & 9 & 7 \
                        hline
                        4 & (B-1) & (B-5)
                        end{array}
                        right.
                        $



                        So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.



                        We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.






                        share|cite|improve this answer


























                          0












                          0








                          0






                          $$2B^2+9B+7mid 7B^2+9B+2$$



                          Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.



                          Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$



                          Writing this out "subtraction-style", we get



                          $left.begin{array}{c}
                          7 & 9 & 2 \
                          2 & 9 & 7 \
                          hline
                          phantom{4}
                          end{array}
                          right.
                          implies
                          left.begin{array}{c}
                          6 & (B+8) & (B+2) \
                          2 & 9 & 7 \
                          hline
                          4 & (B-1) & (B-5)
                          end{array}
                          right.
                          $



                          So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.



                          We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.






                          share|cite|improve this answer














                          $$2B^2+9B+7mid 7B^2+9B+2$$



                          Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.



                          Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$



                          Writing this out "subtraction-style", we get



                          $left.begin{array}{c}
                          7 & 9 & 2 \
                          2 & 9 & 7 \
                          hline
                          phantom{4}
                          end{array}
                          right.
                          implies
                          left.begin{array}{c}
                          6 & (B+8) & (B+2) \
                          2 & 9 & 7 \
                          hline
                          4 & (B-1) & (B-5)
                          end{array}
                          right.
                          $



                          So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.



                          We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 3 hours ago

























                          answered 5 hours ago









                          steven gregory

                          17.7k32257




                          17.7k32257






























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