show that:

for each $n$, there are exsit $n$ consecutive integers, each of which is divisible by the sums of their digits.

Maybe this is old problem,But I can't prove it.can you someone help me? and this simaler problem is link

Thank you for you help

  • $\begingroup$ Can you put an example or two? $\endgroup$ – Manu Manjunath Apr 4 '14 at 13:46
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    $\begingroup$ @Manu for $n=1$ there is $2$ which is the sum of it's digits. $n=2$, $110$ is divisble by $2$, $111$ is divisble by $3$. (That took effort, shamefully) $\endgroup$ – Guy Apr 4 '14 at 13:50
  • $\begingroup$ Yes,Sabyasachi,That's my meaning $\endgroup$ – china math Apr 4 '14 at 13:51
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    $\begingroup$ for $n=3$, $110$ is divisible by $2$, $111$ is divisible by $3$ and $112$ is divisible by $4$ $\endgroup$ – Guy Apr 4 '14 at 13:53
  • $\begingroup$ I think since if it holds for $n$, it obviously holds for all numbers less than $n$, therefore we only have to prove that arbitrarily long stretches can exists. Anybody have any ideas? (although I realize it is trivial for $n\lt10$, just use the single digit numbers). $\endgroup$ – Guy Apr 4 '14 at 13:55

The claim is wrong.

Let $s(x)$ denote the digit sum of $x$. Assume a sequence of $n$ such numbers begins at $m+1$, i.e. $m+k$ is divisible by $s(m+k)$ for $1\le k\le n$. For $n$ big enough (say $n\ge 110$), one of the numbers $m+k$ with $1\le k\le n-10$ ends in the two digits $11$. Then $m+k+9$ ends in $20$, so $s(m+k+9)=s(m+k)$. By assumption, $m+k$ and $m+k+9$ are both divisible by $s(m+k)$. We conclude that $s(m+k)\mid 9$, i.e. $s(m+k)\in\{1,3,9\}$. The same argument works one step further to the right: $m+k+1$ ends in $12$, $m+k+10$ ends in $21$, hence $s(m+k+10)=s(m+k+1)$, and ultimately $s(m+k+1)\in\{1,3,9\}$. But $s(m+k+1)=s(m+k)+1$, contradiction.

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