TAOCP 3.2.1.2 Exercise 1
The modulus $m = 10^{10}$ factors as $2^{10} \cdot 5^{10}$.
Section 3.2.1.2: Choice of Multiplier
Exercise 1. [10] What is the length of the period of the linear congruential sequence with $X_0 = 5772156648$, $a = 3141592621$, $c = 2718281829$, and $m = 10000000000$?
Verified: yes
Solve time: 59s
The modulus $m = 10^{10}$ factors as $2^{10} \cdot 5^{10}$. Let $b = a - 1 = 3141592620$. We check the conditions of Theorem A for a full-period sequence. Condition (i) requires that $\gcd(c, m) = 1$, which is true since $c = 2718281829$ is odd and not divisible by 5. Condition (ii) requires that $b$ be divisible by each prime factor of $m$. Since $b$ is even, it is divisible by 2, but $b \bmod 5 = 3141592620 \bmod 5 = 0$, so it is divisible by 5 as well. Condition (iii) requires that $b$ be divisible by 4 because $m$ is divisible by 4; indeed, $b \bmod 4 = 0$. All three conditions are satisfied. Therefore the linear congruential sequence has period length equal to $m = 10^{10}$.
The period of the sequence is $\boxed{10000000000}$.