nth number which is divisible by all the numbers from 1 to 10












2












$begingroup$


How to find the nth number which is divisible by all the numbers from 1 to 10.



for example



I think 1*2*3*4*5*6*7*8*9*10 = 362880 is the answer for n = 1



Can it's nth term formula be made??






sequences-and-series







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    How to find the nth number which is divisible by all the numbers from 1 to 10.



    for example



    I think 1*2*3*4*5*6*7*8*9*10 = 362880 is the answer for n = 1



    Can it's nth term formula be made??






    sequences-and-series







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      How to find the nth number which is divisible by all the numbers from 1 to 10.



      for example



      I think 1*2*3*4*5*6*7*8*9*10 = 362880 is the answer for n = 1



      Can it's nth term formula be made??






      sequences-and-series







      share|cite|improve this question









      $endgroup$




      How to find the nth number which is divisible by all the numbers from 1 to 10.



      for example



      I think 1*2*3*4*5*6*7*8*9*10 = 362880 is the answer for n = 1



      Can it's nth term formula be made??






      sequences-and-series




      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      sr123sr123

      354




      354






















          2 Answers
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          $begingroup$

          Hint: note that $362880$ has a divisor of $2^7$, while you only need a divisor of $2^3$. There are a number of smaller numbers which are divisible by all the numbers from $1$ to $10$. You need to find the smallest one. Once you find that, any multiple of it will also be divisible by all the numbers from $1$ through $10$, so multiply it by $n$






          share|cite|improve this answer









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            2












            $begingroup$

            I don't think you have that quite right. If something is divisible by 8, it is automatically divisible by 2 and 4, and the same goes for 9 and 3. You want to think about the prime factorisations of all of the numbers from 1-10 and find the lowest common divisor. It would be:



            $$ 5 * 7 * 8 * 9 = 2520 qquad n=1$$



            I'm not sure that this would be the most efficient way to do it. But I'd imagine the next smallest number could be found by multiplying 1260 by any integer you like. So for example, the next number that would satisfy this would be:



            $$ 4 * 5 * 7 * 9 * 2 = 5040 qquad n=2$$
            $$ 4 * 5 * 7 * 9 * 3 = 7560 qquad n=3$$
            $$ 4 * 5 * 7 * 9 * 4 = 10080 qquad n=4$$
            etc...



            So by this logic, I'd say the formula is just:



            $$ T_n = 2520 (n - 1) $$



            I'm not sure if this gets every number. But every number here is guaranteed to be divisible by 10. I'd be interested to see if the answer is different :)






            share|cite|improve this answer









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              $begingroup$

              Hint: note that $362880$ has a divisor of $2^7$, while you only need a divisor of $2^3$. There are a number of smaller numbers which are divisible by all the numbers from $1$ to $10$. You need to find the smallest one. Once you find that, any multiple of it will also be divisible by all the numbers from $1$ through $10$, so multiply it by $n$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Hint: note that $362880$ has a divisor of $2^7$, while you only need a divisor of $2^3$. There are a number of smaller numbers which are divisible by all the numbers from $1$ to $10$. You need to find the smallest one. Once you find that, any multiple of it will also be divisible by all the numbers from $1$ through $10$, so multiply it by $n$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Hint: note that $362880$ has a divisor of $2^7$, while you only need a divisor of $2^3$. There are a number of smaller numbers which are divisible by all the numbers from $1$ to $10$. You need to find the smallest one. Once you find that, any multiple of it will also be divisible by all the numbers from $1$ through $10$, so multiply it by $n$






                  share|cite|improve this answer









                  $endgroup$



                  Hint: note that $362880$ has a divisor of $2^7$, while you only need a divisor of $2^3$. There are a number of smaller numbers which are divisible by all the numbers from $1$ to $10$. You need to find the smallest one. Once you find that, any multiple of it will also be divisible by all the numbers from $1$ through $10$, so multiply it by $n$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Ross MillikanRoss Millikan

                  303k24201375




                  303k24201375























                      2












                      $begingroup$

                      I don't think you have that quite right. If something is divisible by 8, it is automatically divisible by 2 and 4, and the same goes for 9 and 3. You want to think about the prime factorisations of all of the numbers from 1-10 and find the lowest common divisor. It would be:



                      $$ 5 * 7 * 8 * 9 = 2520 qquad n=1$$



                      I'm not sure that this would be the most efficient way to do it. But I'd imagine the next smallest number could be found by multiplying 1260 by any integer you like. So for example, the next number that would satisfy this would be:



                      $$ 4 * 5 * 7 * 9 * 2 = 5040 qquad n=2$$
                      $$ 4 * 5 * 7 * 9 * 3 = 7560 qquad n=3$$
                      $$ 4 * 5 * 7 * 9 * 4 = 10080 qquad n=4$$
                      etc...



                      So by this logic, I'd say the formula is just:



                      $$ T_n = 2520 (n - 1) $$



                      I'm not sure if this gets every number. But every number here is guaranteed to be divisible by 10. I'd be interested to see if the answer is different :)






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        I don't think you have that quite right. If something is divisible by 8, it is automatically divisible by 2 and 4, and the same goes for 9 and 3. You want to think about the prime factorisations of all of the numbers from 1-10 and find the lowest common divisor. It would be:



                        $$ 5 * 7 * 8 * 9 = 2520 qquad n=1$$



                        I'm not sure that this would be the most efficient way to do it. But I'd imagine the next smallest number could be found by multiplying 1260 by any integer you like. So for example, the next number that would satisfy this would be:



                        $$ 4 * 5 * 7 * 9 * 2 = 5040 qquad n=2$$
                        $$ 4 * 5 * 7 * 9 * 3 = 7560 qquad n=3$$
                        $$ 4 * 5 * 7 * 9 * 4 = 10080 qquad n=4$$
                        etc...



                        So by this logic, I'd say the formula is just:



                        $$ T_n = 2520 (n - 1) $$



                        I'm not sure if this gets every number. But every number here is guaranteed to be divisible by 10. I'd be interested to see if the answer is different :)






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          I don't think you have that quite right. If something is divisible by 8, it is automatically divisible by 2 and 4, and the same goes for 9 and 3. You want to think about the prime factorisations of all of the numbers from 1-10 and find the lowest common divisor. It would be:



                          $$ 5 * 7 * 8 * 9 = 2520 qquad n=1$$



                          I'm not sure that this would be the most efficient way to do it. But I'd imagine the next smallest number could be found by multiplying 1260 by any integer you like. So for example, the next number that would satisfy this would be:



                          $$ 4 * 5 * 7 * 9 * 2 = 5040 qquad n=2$$
                          $$ 4 * 5 * 7 * 9 * 3 = 7560 qquad n=3$$
                          $$ 4 * 5 * 7 * 9 * 4 = 10080 qquad n=4$$
                          etc...



                          So by this logic, I'd say the formula is just:



                          $$ T_n = 2520 (n - 1) $$



                          I'm not sure if this gets every number. But every number here is guaranteed to be divisible by 10. I'd be interested to see if the answer is different :)






                          share|cite|improve this answer









                          $endgroup$



                          I don't think you have that quite right. If something is divisible by 8, it is automatically divisible by 2 and 4, and the same goes for 9 and 3. You want to think about the prime factorisations of all of the numbers from 1-10 and find the lowest common divisor. It would be:



                          $$ 5 * 7 * 8 * 9 = 2520 qquad n=1$$



                          I'm not sure that this would be the most efficient way to do it. But I'd imagine the next smallest number could be found by multiplying 1260 by any integer you like. So for example, the next number that would satisfy this would be:



                          $$ 4 * 5 * 7 * 9 * 2 = 5040 qquad n=2$$
                          $$ 4 * 5 * 7 * 9 * 3 = 7560 qquad n=3$$
                          $$ 4 * 5 * 7 * 9 * 4 = 10080 qquad n=4$$
                          etc...



                          So by this logic, I'd say the formula is just:



                          $$ T_n = 2520 (n - 1) $$



                          I'm not sure if this gets every number. But every number here is guaranteed to be divisible by 10. I'd be interested to see if the answer is different :)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          user2662833user2662833

                          1,098815




                          1,098815






























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