jasperE3 wrote: For any $n\in\mathbb N$, denote by $a_n$ the sum $2+22+222+\cdots+22\ldots2$, where the last summand consists of $n$ digits of $2$. Determine the greatest $n$ for which $a_n$ contains exactly $222$ digits of $2$. We have $a_1=2$ and $a_{n+1}=10a_n+2n+2$ And so $a_{n+1}+\frac 29(n+1)+\frac{20}{81}=10(a_n+\frac 29n+\frac{20}{81}$ And so $a_n+\frac 29n+\frac{20}{81}=10^{n-1}(a_1+\frac 291+\frac{20}{81})=\frac 2{81}10^{n+1}$ And so $a_n=\frac{2\times 10^{n+1}-18n-20}{81}$ And so $\boxed{a_n=\left\lfloor\frac 2{81}10^{n+1}\right\rfloor-\left\lfloor\frac{18n+20}{81}\right\rfloor}$ Note then that $\frac 2{81}=0.\overline{024691358}...$ and so : $a_{9k-1}=\overbrace{024691358}^{k\times}-2k$ $a_{9k}=\overbrace{024691358}^{k\times}-2k$ $a_{9k}=\overbrace{024691358}^{k\times}0-2k$ $a_{9k+1}=\overbrace{024691358}^{k\times}02-2k$ $a_{9k+2}=\overbrace{024691358}^{k\times}024-2k$ $a_{9k+3}=\overbrace{024691358}^{k\times}0246-2k$ $a_{9k+4}=\overbrace{024691358}^{k\times}02469-2k-1$ $a_{9k+5}=\overbrace{024691358}^{k\times}024691-2k-1$ $a_{9k+6}=\overbrace{024691358}^{k\times}0246913-2k-1$ $a_{9k+7}=\overbrace{024691358}^{k\times}02469135-2k-1$ and so obviously at least $223$ digits $2$ if $k\ge 223$ in each cases Testing $k=222$ $a_{9\times 222-1}=\overbrace{024691358}^{222\times}-444$ $=\overbrace{024691358}^{221\times}024690914$ $a_{9\times 222}=\overbrace{024691358}^{222\times}0-444$ $=\overbrace{024691358}^{221\times}0246913136$ $a_{9\times 222+1}=\overbrace{024691358}^{222\times}02-444$ $=\overbrace{024691358}^{221\times}02469135358$ $a_{9\times 222+2}=\overbrace{024691358}^{222\times}024-444$ $=\overbrace{024691358}^{221\times}024691357580$ $a_{9\times 222+3}=\overbrace{024691358}^{222\times}0246-444$ $=\overbrace{024691358}^{221\times}0246913579802$ $a_{9\times 222+4}=\overbrace{024691358}^{222\times}02469-445$ $=\overbrace{024691358}^{222\times}02024$ $a_{9\times 222+5}=\overbrace{024691358}^{222\times}024691-445$ $=\overbrace{024691358}^{222\times}024246$ $a_{9\times 222+6}=\overbrace{024691358}^{222\times}0246913-445$ $=\overbrace{024691358}^{222\times}0246468$ $a_{9\times 222+7}=\overbrace{024691358}^{222\times}02469135-445$ $=\overbrace{024691358}^{222\times}02468690$ Hence the answer $\boxed{a_{9\times 222+2}=a_{2000}}$