THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Let $\overline{a_1{a}_2\dots a_{n-2}{a}_{n-1}{a}_n}$ be the number we search for, so $$\overline{a_1{a}_2\dots a_{n-2}{a}_{n-1}{a}_n}-2017=\overline{a_1{a}_2\dots a_{n-2}},$$ which is equivalent to $\overline{a_1{a}_2\dots a_{n-2}{a}_{n-1}{a}_n}-\overline{a_1{a}_2\dots a_{n-2}}=2017$. On the other hand, $$\overline{a_1{a}_2\dots a_{n-2}{a}_{n-1}{a}_n}-\overline{a_1{a}_2\dots a_{n-2}}>10^{n-1}-10^{n-2}=9\cdot 10^{n-2},$$ hence $2017>9\cdot 10^{n-2}$, and it follows that $n-2\le 2$, or $n\le 4$, and since our number is greater than 2017, it has precisely four digits. $$\overline{abcd}=2017+\overline{ab}\le 2017+99=2116\text{ and }\overline{abcd}\ge 2017\text{ imply that }a=2.$$ $$\text{Thus, }2bcd=2017+2b\le 2017+29=2046,\text{ so }b=0.$$ Consequently, the number we searched is $2017+20=2037$.