AIME 2026 II · 第 9 题

Solution 1

Expressing $S$ algebraically, we can write

$$S=\sum_{a\ge1}\frac{1}{10^a-1}=\sum_{a\ge1}\sum_{b\ge1}10^{-ab}.$$ The term $10^{-n}$ will appear exactly $\tau(n)$ times in the above sum: this is because the divisor function $\tau(n)$ counts the number of ordered pairs $(a,b)$ with $a,b\ge 1$ and $ab=n$. So we have

$$S=\sum_{n\ge 1}\tau(n)10^{-n},\quad10^{100}S = \sum_{n\ge 1}\tau(n)10^{100-n}.$$ The sum of all terms with $n>100$ is less than $1$, so it is negligible. Hence

$$\lfloor 10^{100}S\rfloor \equiv \sum_{n=1}^{100}\tau(n)10^{100-n}\equiv\tau(98)\cdot 100+\tau(99)\cdot 10+\tau(100)\equiv \boxed{669} \pmod{1000}.$$ ~TThB0501

Solution 2 (Factors)

After a bit of thinking, you'll realize that for each digit after $0.\dots$ is the number of factors of that digit.

So we're basically looking for the number of factors of $98,99,100 ($and $101$ to check for carried ones$).$

Factors of $98 = 2\cdot7^2, 99 = 11\cdot3^2, 100 = 2^2\cdot5^2, 101 = 101, 98 \to{2\cdot3=6}, 99 \to{2\cdot3=6}, 100 \to{3\cdot3=9}, 101 \to{2}.$

So, we get $669$ and $2$. The two doesn't carry a one, therefore the answer is $\boxed{669}.$

~ BenjaminDong01

Solution 3 (Decimals)

We can represent each of the terms as a recurring decimal.

For example, $\frac{1}{9} = 0.\overline{1}$ and $\frac{1}{99} = 0.\overline{01}$. Given this it is trivial to see that the only terms that will contribute to the $nth$ decimal place will be $\frac{1}{10^a - 1}$ where $a$ is a factor of $n$.

Now consider multiplying $S$ by $10^{100}$ and taking it modulo $1000$. This would leave the digits in the 98th, 99th, and 100th places.

So, we find that 98 has 6 factors, 99 has 6 factors, and 100 has 9 factors. 101 is prime, so we don't need to worry about any carrying affecting the 100th decimal place.

Therefore, the answer is $\boxed{669}$.

~mathmaniak

Solution 4 (Really thorough)

It is very well known that $\frac{a}{10^n-1}=0.\overline{\underbrace{00...00}_{\text{ n-1 zeros}}a}$.

Therefore, we can say that

$$T_n=\frac{1}{10^n-1}=0.\overline{\underbrace{00...00}_{\text{ n-1 zeros}}1}$$ .

Let $S_k=\sum_{n=0}^k(T_n)$,

The problem asks us for

$$\lfloor10^{100}S_\infty\rfloor\pmod{1000}$$ .

let $D_n$ be the $n-th$ place value of $S_\infty$, before converting. For example, $D_{48}=10$. Observe that each $T_n$ is periodic with period $n$. Then, it sbould be obvious that each value of $D_{kn}$ ($k\in\mathbb{Z^+}$) has a $1$ contributed from $T_n$, as $T_n$ "deposits" as $1$ every $n$ digits.

We can therefore say that $D_n=\sigma_0(n)$ ($\sigma_0(n)$ is the number of divisors of n. In general, $\sigma_k(n)$ is the sum of the $k-th$ powers of the divisors n)

I think we can agree that $S_\infty=\sum_{n=0}^\infty(\frac{D^n}{10^n})=\sum_{n=0}^\infty(\frac{\sigma_0(n)}{10^n})$ from the definition of place values.

We can also argue that the terms after $D_{100}$ are inconsequential. This is basically because it is easy enough to see that $\sigma_0(n)$ is much less than $10^{n-100}$ if $n>100$, and summing all of those values is logically not going to be over 1. IE, those terms will never "escape the decimal place". Then, the floor part will just remove that.

So we can say that

$$\lfloor10^{100}S_\infty\rfloor=10^{100}S_{100}=10^{100}\sum_{n=0}^{100}(\frac{\sigma_0(n)}{10^n})$$ We want:

$$10^{100}\sum_{n=0}^{100}(\frac{\sigma_0(n)}{10^n})\pmod{1000}$$ Since we are working mod 1000, all of the terms with $n\leq97$ will end up as $\sigma_0(n)*10^{(r\geq3)}$. So, all of the terms just disappear before n=98. We are then left with:

$${\sigma_0(98)}*{10^{(100-98)}}+{\sigma_0(99)}*{10^{(100-99)}}+{\sigma_0(100)}*{10^{(100-100)}}=6*100+6*10+9=\boxed{669}$$ Because $98=2*49=2^1*7^2\therefore\sigma_0(98)=2*3=6$ and $99=9*11=3^2*11^2\therefore\sigma_0(99)=2*3=6$ and $100=4*25=2^2*5^2\therefore\sigma_0(100)=3*3=9$

~Stereotypicalmathnerd

Solution 5 (Intuition for Solution 1)

Like in solution 1, we see that we just have the sum $\sum_{k=1}^{\infty} \frac{1}{10^k - 1}$. We factor out $10^k$ at the bottom to obtain $\sum_{k=1}^{\infty} \frac{1}{10^k(1 - 10^{-k})}$. We replace $\frac{1}{10^k}$ as $10^{-k}$ and the result is $\sum_{k=1}^{\infty} \frac{10^{-k}}{1 - 10^{-k}}$. However, that result is just an infinite geometric series! We have that $r = 10^{-k}$ and $0 < |r| < 1$. The first term of this series is $r$ itself. Thus, the series is $r^l$ for some values $l$, and we can conclude the sum as $\sum_{l=1}^{\infty} (10^{-k})^l$. We determine the value for $k$ using an exterior nested sum to obtain the double sum found in solution 1. Continue as followed.

~Pinotation