NUMBER system and counting

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Let $S =1,2,3,...,20$ be the set of all positive from $1$ to $20$ suppose that $N$ is the smallest positive integer such that exactly eighteen numbers from $S$ are factors of $N$ and the only two numbers from $S$ that are not factors of $N$ are consecutive integers. Find the sum of digits of $N$.



We first find out which two consecutive numbers from $S$ are
not factors of $N$. Clearly $1$ is the factor of $N$.
If $k$ is not factor of $N$ then $2k$ will also be not he factor of $N$.



How will I solve further please help.



Thanks




number-theory number-systems




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    Let $S =1,2,3,...,20$ be the set of all positive from $1$ to $20$ suppose that $N$ is the smallest positive integer such that exactly eighteen numbers from $S$ are factors of $N$ and the only two numbers from $S$ that are not factors of $N$ are consecutive integers. Find the sum of digits of $N$.



    We first find out which two consecutive numbers from $S$ are
    not factors of $N$. Clearly $1$ is the factor of $N$.
    If $k$ is not factor of $N$ then $2k$ will also be not he factor of $N$.



    How will I solve further please help.



    Thanks




    number-theory number-systems




    share|cite|improve this question
























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      Let $S =1,2,3,...,20$ be the set of all positive from $1$ to $20$ suppose that $N$ is the smallest positive integer such that exactly eighteen numbers from $S$ are factors of $N$ and the only two numbers from $S$ that are not factors of $N$ are consecutive integers. Find the sum of digits of $N$.



      We first find out which two consecutive numbers from $S$ are
      not factors of $N$. Clearly $1$ is the factor of $N$.
      If $k$ is not factor of $N$ then $2k$ will also be not he factor of $N$.



      How will I solve further please help.



      Thanks




      number-theory number-systems




      share|cite|improve this question














      Let $S =1,2,3,...,20$ be the set of all positive from $1$ to $20$ suppose that $N$ is the smallest positive integer such that exactly eighteen numbers from $S$ are factors of $N$ and the only two numbers from $S$ that are not factors of $N$ are consecutive integers. Find the sum of digits of $N$.



      We first find out which two consecutive numbers from $S$ are
      not factors of $N$. Clearly $1$ is the factor of $N$.
      If $k$ is not factor of $N$ then $2k$ will also be not he factor of $N$.



      How will I solve further please help.



      Thanks





      number-theory number-systems





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      share|cite|improve this question




      share|cite|improve this question








      edited Aug 18 at 5:05









      apanpapan3

      1231211




      1231211










      asked Aug 18 at 4:59









      Mamta Gupta

      63




      63




















          1 Answer
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          Hint: start down the possible pairs of nonfactors. It can't be $19,20$ because then we know $4$ and $5$ are both factors, so $20$ must be. Now try $18,19$, which fails (why?). Keep going.






          share|cite|improve this answer




















          • Yes I got it then 16 and 17 are not the factors of N
            – Mamta Gupta
            Aug 18 at 5:10











          • Correct. Can you find the prime decomposition of $N$? Where are you stuck?
            – Ross Millikan
            Aug 18 at 5:19











          • Yes 2^3×3^2×5×7×11×13×19=6846840
            – Mamta Gupta
            Aug 18 at 5:31










          • Is it correct....
            – Mamta Gupta
            Aug 18 at 5:31






          • 2




            Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
            – Ross Millikan
            Aug 18 at 5:35










          Your Answer




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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          up vote
          1
          down vote













          Hint: start down the possible pairs of nonfactors. It can't be $19,20$ because then we know $4$ and $5$ are both factors, so $20$ must be. Now try $18,19$, which fails (why?). Keep going.






          share|cite|improve this answer




















          • Yes I got it then 16 and 17 are not the factors of N
            – Mamta Gupta
            Aug 18 at 5:10











          • Correct. Can you find the prime decomposition of $N$? Where are you stuck?
            – Ross Millikan
            Aug 18 at 5:19











          • Yes 2^3×3^2×5×7×11×13×19=6846840
            – Mamta Gupta
            Aug 18 at 5:31










          • Is it correct....
            – Mamta Gupta
            Aug 18 at 5:31






          • 2




            Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
            – Ross Millikan
            Aug 18 at 5:35














          up vote
          1
          down vote













          Hint: start down the possible pairs of nonfactors. It can't be $19,20$ because then we know $4$ and $5$ are both factors, so $20$ must be. Now try $18,19$, which fails (why?). Keep going.






          share|cite|improve this answer




















          • Yes I got it then 16 and 17 are not the factors of N
            – Mamta Gupta
            Aug 18 at 5:10











          • Correct. Can you find the prime decomposition of $N$? Where are you stuck?
            – Ross Millikan
            Aug 18 at 5:19











          • Yes 2^3×3^2×5×7×11×13×19=6846840
            – Mamta Gupta
            Aug 18 at 5:31










          • Is it correct....
            – Mamta Gupta
            Aug 18 at 5:31






          • 2




            Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
            – Ross Millikan
            Aug 18 at 5:35












          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint: start down the possible pairs of nonfactors. It can't be $19,20$ because then we know $4$ and $5$ are both factors, so $20$ must be. Now try $18,19$, which fails (why?). Keep going.






          share|cite|improve this answer












          Hint: start down the possible pairs of nonfactors. It can't be $19,20$ because then we know $4$ and $5$ are both factors, so $20$ must be. Now try $18,19$, which fails (why?). Keep going.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 18 at 5:06









          Ross Millikan

          278k21187354




          278k21187354











          • Yes I got it then 16 and 17 are not the factors of N
            – Mamta Gupta
            Aug 18 at 5:10











          • Correct. Can you find the prime decomposition of $N$? Where are you stuck?
            – Ross Millikan
            Aug 18 at 5:19











          • Yes 2^3×3^2×5×7×11×13×19=6846840
            – Mamta Gupta
            Aug 18 at 5:31










          • Is it correct....
            – Mamta Gupta
            Aug 18 at 5:31






          • 2




            Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
            – Ross Millikan
            Aug 18 at 5:35
















          • Yes I got it then 16 and 17 are not the factors of N
            – Mamta Gupta
            Aug 18 at 5:10











          • Correct. Can you find the prime decomposition of $N$? Where are you stuck?
            – Ross Millikan
            Aug 18 at 5:19











          • Yes 2^3×3^2×5×7×11×13×19=6846840
            – Mamta Gupta
            Aug 18 at 5:31










          • Is it correct....
            – Mamta Gupta
            Aug 18 at 5:31






          • 2




            Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
            – Ross Millikan
            Aug 18 at 5:35















          Yes I got it then 16 and 17 are not the factors of N
          – Mamta Gupta
          Aug 18 at 5:10





          Yes I got it then 16 and 17 are not the factors of N
          – Mamta Gupta
          Aug 18 at 5:10













          Correct. Can you find the prime decomposition of $N$? Where are you stuck?
          – Ross Millikan
          Aug 18 at 5:19





          Correct. Can you find the prime decomposition of $N$? Where are you stuck?
          – Ross Millikan
          Aug 18 at 5:19













          Yes 2^3×3^2×5×7×11×13×19=6846840
          – Mamta Gupta
          Aug 18 at 5:31




          Yes 2^3×3^2×5×7×11×13×19=6846840
          – Mamta Gupta
          Aug 18 at 5:31












          Is it correct....
          – Mamta Gupta
          Aug 18 at 5:31




          Is it correct....
          – Mamta Gupta
          Aug 18 at 5:31




          2




          2




          Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
          – Ross Millikan
          Aug 18 at 5:35




          Yes, that is correct. Once you have $N$ you have the sum of digits. It is interesting to try to state the condition on the pair of missing numbers based on the maximum number in $S$.
          – Ross Millikan
          Aug 18 at 5:35












           

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