Question

The fraction $\dfrac{1}{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},$ where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots+b_{n-1}?$

$874$

$883$

$887$

$891$

$892$

Solution:

Reading the block in pairs of digits (base $100$ ), $\dfrac{1}{9801}=\dfrac{1}{99^2}$ expands as $00,01,02,\ldots,$ since $\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}.$ Carrying works out so that every two-digit block $00,01,\ldots,97$ appears, the block $98$ is skipped, and $99$ appears, before the period repeats.

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.

2014 AMC 12A Problem 23

Solution:

Problem 23 in Other Years