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.



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.

  • $\begingroup$ Yes I got it then 16 and 17 are not the factors of N $\endgroup$ – Mamta Gupta Aug 18 '18 at 5:10
  • $\begingroup$ Correct. Can you find the prime decomposition of $N$? Where are you stuck? $\endgroup$ – Ross Millikan Aug 18 '18 at 5:19
  • $\begingroup$ Yes 2^3×3^2×5×7×11×13×19=6846840 $\endgroup$ – Mamta Gupta Aug 18 '18 at 5:31
  • $\begingroup$ Is it correct.... $\endgroup$ – Mamta Gupta Aug 18 '18 at 5:31
  • 2
    $\begingroup$ 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 $S$. $\endgroup$ – Ross Millikan Aug 18 '18 at 5:35

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