Instant trick

Note that $9901 = 99\times100 + 1 = 100^2 - 100 + 1 \; = \; 10^4 - 10^2 + 1$ , so that $101 \times 9901 \; = \; (10^2 + 1)(10^4 - 10^2 + 1) \; = \; 10^6 + 1$ and hence $9901$ divides $10^{12}-1 = 9 \times (1)^{12}$ .

If $k$ is the smallest positive integer such that $9901$ divides $(1)^k = \tfrac19(10^k - 1)$ , then $k$ is the smallest positive integer such that $10^k \equiv 1 \pmod{9901}$ , and hence $k$ divides $12$ . But $10^6 \equiv -1 \pmod{9901}$ and $10^4 \equiv 99 \pmod{9901}$ (see the identities in the first paragraph, while $10,10^2,10^3$ are all smaller than $9901$ . Thus we deduce that the required answer is $\boxed{12}$ .

All one needs to do is to divide $1$ by $9901$ in the calculator. The fraction is rational and it is of the repeating decimal part type. The number of decimal points, before the repeating starts, would be the answer.

To understand the mechanics, one should know that $0.\overline{a_{n-1}...a_0}=\frac{\overline{a_{n-1}...a_0}}{9 \times (1)^n}=\frac{A}{B}$ , where $gcd(A,B)=1$ .

Then, there exists $k \in \mathbb{N}$

$\frac{1}{9901}=0.\overline{a_{k-1}...a_0}=\frac{\overline{a_{k-1}...a_0}}{9 \times (1)^k} \implies 9901\times \overline{a_{n-1}...a_0}=9 \times (1)^k \implies 9901 | (1)^k$

and, clearly, $k$ is the number of the decimals of $0.\overline{a_{k-1}...a_0}$ .

calculate 1/9901 in the calculator and find the length of the repeating units after decimal.