2014 AMC 12A Problem 23

Since $\dfrac{1}{99^2}=\dfrac{1}{9801}$, the repeating block satisfies $10^n-1=9801\cdot\overline{b_{n-1}\ldots b_0}$

Group the repeating decimal into base-$100$ digits and use $(100-1)^{-2};$ follow the carries until the first base-$100$ digit repeats

Reading the block in pairs of digits (base $100$), $\dfrac{1}{9801}=\dfrac{1}{99^2}$ expands as $00,01,02,\ldots,$ since $\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}.$ Let $a_j$ be the $j$th base-$100$ digit of the repeating block. Multiplying the block by $99^2$ shows first that $a_0=99.$ The resulting carry gives $a_1=97,$ after which there is no carry and successively $a_j=98-j$ for $1\le j\le98.$ The next digit is again $99,$ so the period is $00,01,02,\ldots,96,97,99,$ with $98$ omitted.

If the blocks $00$ through $99$ all appeared, the digit sum would be $(0+1+\cdots+9)\cdot20=900.$ Removing the missing $98$ subtracts $9+8,$ giving $900-9-8=883.$

Thus, the correct answer is B.