2026 AIME II Problem 9

Question

Let

S

denote the value of the infinite sum

\frac{1}{9} + \frac{1}{99} + \frac{1}{999} + \frac{1}{9999} + \cdots

Find the remainder when the greatest integer less than or equal to

10^{100} S

is divided by

1000.

Answer

Expand each

\frac{1}{10^k - 1}

as a geometric series in

10^{-k}

and collect equal powers of

10:

the coefficient of

10^{-n}

counts the divisors of

n.

After multiplying by

10^{100},

only

n = 98, 99, 100

affect the last three digits, and the leftover tail is between

0

and

1.

Each term is

\frac{1}{10^k - 1} = \sum_{j \ge 1} 10^{-kj},

so summing over

k

and collecting the exponent

n = kj,

S = \sum_{n \ge 1} \frac{d(n)}{10^n},

where

d(n)

is the number of divisors of

n.

Hence

10^{100} S = \sum_{n = 1}^{100} d(n)\,10^{100 - n}

+ T

with

T = \sum_{m \ge 1} d(100 + m)\,10^{-m}.

From

d(101) = 2,

d(102) = 8,

d(103) = 2,

d(104) = 8,

the tail starts

0.2

+ 0.08

+ 0.002

+ 0.0008 = 0.2828,

and since

d(N) \lt 2\sqrt{N},

the remaining terms contribute less than

\sum_{m \ge 5} \frac{2\sqrt{100 + m}}{10^m} \lt 0.001.

So

0 \lt T \lt 1

and

\left\lfloor 10^{100} S \right\rfloor = \sum_{n = 1}^{100} d(n)\,10^{100 - n}.

Modulo

1000,

every term with

n \le 97

is a multiple of

1000,

leaving

d(98) \cdot 100 + d(99) \cdot 10

+ d(100).

Since

d(98) = 6,

d(99) = 6,

and

d(100) = 9,

the remainder is

600 + 60 + 9 = 669.

Problem 9 in other years