How to find the unknown number while only LCM is given












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Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
I have searched many ways to solve but could not find a step wise method to teach the students, please help.










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    0












    $begingroup$


    Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
    I have searched many ways to solve but could not find a step wise method to teach the students, please help.










    share|cite|improve this question











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      0












      0








      0





      $begingroup$


      Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
      I have searched many ways to solve but could not find a step wise method to teach the students, please help.










      share|cite|improve this question











      $endgroup$




      Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
      I have searched many ways to solve but could not find a step wise method to teach the students, please help.







      least-common-multiple






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      edited Mar 5 '18 at 17:48









      rbird

      1,194114




      1,194114










      asked Mar 5 '18 at 17:41









      Yunas KhanYunas Khan

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          4 Answers
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          $begingroup$

          $15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.






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            Hint:   $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.






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              $begingroup$

              I would present the following as a systematic approach
              $$
              eqalign{
              & {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
              & Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
              $$






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                $begingroup$

                $n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
                $$1,3,5,9,15,45.$$
                Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.






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                  4 Answers
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                  4 Answers
                  4






                  active

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                  active

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                  active

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                  0












                  $begingroup$

                  $15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.






                  share|cite|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    $15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.






                    share|cite|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      $15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.






                      share|cite|improve this answer









                      $endgroup$



                      $15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Mar 5 '18 at 17:43









                      rbirdrbird

                      1,194114




                      1,194114























                          0












                          $begingroup$

                          Hint:   $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Hint:   $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Hint:   $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.






                              share|cite|improve this answer









                              $endgroup$



                              Hint:   $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.







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                              share|cite|improve this answer










                              answered Mar 5 '18 at 17:44









                              dxivdxiv

                              58k648102




                              58k648102























                                  0












                                  $begingroup$

                                  I would present the following as a systematic approach
                                  $$
                                  eqalign{
                                  & {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
                                  & Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    I would present the following as a systematic approach
                                    $$
                                    eqalign{
                                    & {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
                                    & Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      I would present the following as a systematic approach
                                      $$
                                      eqalign{
                                      & {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
                                      & Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
                                      $$






                                      share|cite|improve this answer









                                      $endgroup$



                                      I would present the following as a systematic approach
                                      $$
                                      eqalign{
                                      & {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
                                      & Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
                                      $$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Mar 5 '18 at 18:06









                                      G CabG Cab

                                      20.4k31341




                                      20.4k31341























                                          0












                                          $begingroup$

                                          $n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
                                          $$1,3,5,9,15,45.$$
                                          Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            $n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
                                            $$1,3,5,9,15,45.$$
                                            Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              $n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
                                              $$1,3,5,9,15,45.$$
                                              Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.






                                              share|cite|improve this answer









                                              $endgroup$



                                              $n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
                                              $$1,3,5,9,15,45.$$
                                              Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Mar 5 '18 at 18:52









                                              farruhotafarruhota

                                              21.3k2841




                                              21.3k2841






























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