Japan Junior Mathematical Olympiad

A positive integer $n$ less than or equal to $999$ can be represented as $10^{2}a+10b+c$ where $a,b,c$ are integers satisfying $0\le a,b,c\le 9$ and $a+b+c\ne 0$. Then we have $S(n)=a+b+c$ and $$S(n+1)=\{\begin{array}{ll}a+b+c+1,& \text{if }c<9\cdots (\text{i}),\\ a+b+1,& \text{if }c=9,b<9\cdots (\text{ii}),\\ a+1,& \text{if }c=b=9,a<9\cdots (\text{iii}),\\ 1,& \text{if }c=b=a=9\cdots (\text{iv}).\end{array}$$

Now, it is clear that $\frac{S(n)}{S(n+1)}$ is not an integer in case (i), and is the integer $999$ in case (iv) above. In case (iii), we have $\frac{S(n)}{S(n+1)}=\frac{a+18}{a+1}$, which is an integer only when $a=0$, corresponding to the case $n=99$. Finally, in case (ii), $\frac{S(n)}{S(n+1)}=\frac{a+b+9}{a+b+1}$, with $0\le a\le 9$, $0\le b<9$. If we let $k=a+b$, then we see that $\frac{k+9}{k+1}$ is an integer only when $k=0,1,3,7$, from which we conclude that only possibilities for $(a,b)$ under the case (ii) are $$(a,b)=(0,0),(0,1),(0,3),(0,7),(1,0),(1,2),(1,6),(2,1),(2,5),(3,0),(3,4),(4,3),(5,2),(6,1),(7,0),$$ and these choices of $(a,b)$ correspond to $$n=9,19,39,79,109,129,169,219,259,309,349,439,529,619,709,$$ respectively. These, together with the numbers $99$ and $999$ found above, give $17$ integers satisfying the condition of the problem.