TAOCP 3.3.3 Exercise 4

Section 3.3.3: Theoretical Tests

Exercise 4. ▶ [M19] If $m = 10^{10}$, what is the highest possible value of $d$ (in the notation of Theorem P), given that the potency of the generator is 10?

Verified: yes
Solve time: 2m01s

Solution

The problem asks for the maximum possible value of $d$ in the notation of Theorem P, given that $m = 10^{10}$ and the potency of the generator is 10. We recall that in Section 3.3.3, Theorem P introduces the parameter $d = \gcd(b, m)$, where $b = a - 1$, and that the potency of a linear congruential generator with modulus $m$ and multiplier $a$ is defined as the maximal integer $t$ such that $d^t$ divides $m$. In other words, the potency of the generator is the largest exponent $t$ for which $d^t \mid m$.

Here, we are told that the potency is 10 and $m = 10^{10}$. Factor $m$:

$$ m = 10^{10} = 2^{10} \cdot 5^{10}. $$

Let $d$ be the greatest common divisor of $b = a-1$ and $m$, written in the prime factorization form:

$$ d = 2^u \cdot 5^v, \quad 0 \le u, v \le 10. $$

The potency $t$ is then

$$ t = \min{ 10/u, 10/v }, $$

where we interpret $10/0$ as infinity if one of the exponents is zero. The potency is defined as the largest $t$ such that $d^t \mid m$, equivalently:

$$ d^{10} \le m \quad \text{with equality if $d$ is maximal}. $$

Since $m = 2^{10} 5^{10}$, any $d = 2^u 5^v$ raised to the 10th power must divide $m$:

$$ d^{10} = 2^{10 u} \cdot 5^{10 v} \mid 2^{10} \cdot 5^{10}. $$

This requires

$$ 10 u \le 10 \implies u \le 1, \quad 10 v \le 10 \implies v \le 1. $$

Thus the largest possible $d$ consistent with potency 10 is

$$ d = 2^1 \cdot 5^1 = 10. $$

No larger value is possible, because increasing either $u$ or $v$ to 2 would make $d^{10} = 2^{20} \cdot 5^{10}$ or $2^{10} \cdot 5^{20}$, which exceeds $m$ and violates the potency condition.

Hence the highest possible value of $d$ is

$$ \boxed{10}. $$

This completes the proof. ∎