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 $b_0+b_1+\cdots+b_{n-1}?$

$874$

$883$

$887$

$891$

$892$

Answer: B

Concepts:repeating decimal digits pattern recognition

Difficulty rating: 2380

Solution:

Reading the block in pairs of digits (base $100$ ), $\dfrac{1}{9801}=\dfrac{1}{99^2}$ expands as $00,01,02,\ldots,$ since $\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}.$ 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.

2014 AMC 12A Problem 23

Answer: B

Difficulty rating: 2380

Solution:

Problem 23 in Other Years