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A repeating digit records a remainder state, but it need not identify that state. For a prime p not dividing a positional base b, we prove that the digit map is injective on the nonzero remainders exactly when p\le b+1. In this range, two synchronized periodic tails either agree at every position or disagree at every position, even when the remainders belong to several cycles.
Let n=pm, where every prime factor of m divides b. Compare all proper fractions k/n with 1/n after one common depth that clears m, giving each terminating fraction score one. For every digit-partitioning prime, the mean alignment is \alpha_b(pm)=\frac{2m-1}{pm-1}. The cycle lengths and the number of cycles disappear from this count. In base ten, the same formula covers the fixed repetends at 3, the full cycle at 7, and the complement pairs at 11. The boundary p\le b+1 identifies exactly when digit agreement recovers equality of remainder states and the alignment reduces to counting complete matches.
In base ten, the ten proper fractions k/11 use every decimal digit exactly once at each position of their repetends. The twelve proper fractions k/13 cannot do this. Their leading digits repeat 3 and 6, since twelve nonzero remainder states must enter ten digit bins. At 11, observing a digit identifies its remainder. At 13, a shared digit can come from two different states.
This change determines how the fraction table can be counted. Long division advances remainders by a permutation. If different states write different digits at one position, they remain separate at every position. Synchronized periodic tails therefore give complete agreement or none. Once distinct states can write the same digit, partial agreement becomes possible.
For every prime p\nmid b, the exact boundary is p\le b+1. Let n=pm with m supported on the prime factors of b, and compare every proper fraction with 1/n after a common division depth that clears m. In the injective range, the m-1 terminating rows receive score one, exactly m nonterminating rows match throughout, and every other row disagrees throughout. Thus \alpha_b(pm)=\frac{2m-1}{pm-1}. The decimal cases p=3, p=7, and p=11 have fixed states, one cycle of six, and five complement pairs respectively. Their repeating patterns differ, but the alignment requires the same count. The lengths and arrangement of the cycles are unnecessary once the digit map distinguishes their states.
Coordinatewise agreement under cyclic shift is classical Hamming correlation [1]. Kak and Chatterjee study this observable for full-period reciprocal digit sequences, including complement structure and Hamming-distance and autocorrelation bounds [2]. Repetend multiplication, varying bases, and Midy-type complement phenomena are also established [3, 4]. The sharp digit boundary and the resulting count concern the complete denominator table. They do not require the base to generate the unit group modulo p.
The one-digit alignment at prime 3 [5] is one case of this common mechanism. The limit 2/p selects 3 as the only prime p\ge3 in the digit-partitioning class that can cross the reciprocal golden threshold. The case is available whenever 3\nmid b. The injectivity boundary establishes how far the same count applies before distinct remainder states can contribute partial matches.
Let b \ge 2 be a positional base and p a prime not dividing b. The multiplicative order L = \mathop{\mathrm{ord}}_p(b) is the length of the repetend of 1/p in base b [8]. The group (\mathbb{Z}/p\mathbb{Z})^{*} has order p - 1, and the cyclic subgroup \langle b \rangle has order L, partitioning \{1, \ldots, p-1\} into (p-1)/L cosets. Within each coset, the repetends of k/p are cyclic permutations of a common string.
Definition 1 (Base-supported integers). A positive integer m is b-supported if every prime factor of m divides b. Equivalently, m divides some power of b.
Definition 2 (Synchronized repetend alignment). For n = pm with m a b-supported integer, the repetend alignment \alpha_b(n) is defined at a common division depth. Choose t\ge0 with m\mid b^t. Beginning immediately after the first t base-b digits, compare the next L=\mathop{\mathrm{ord}}_p(b) digits of k/n with the corresponding digits of 1/n. For a terminating fraction, set a_b(k/n,1/n)=1. Otherwise, let a_b(k/n,1/n) be the fraction of the L synchronized positions at which the two digits agree. Then \alpha_b(n) is the mean of these scores over k \in \{1, \ldots, n-1\}.
After depth t, every nonterminating tail has reduced denominator p and period L. The definition does not depend on the choice of t. Replacing t by a larger clearing depth advances both nonterminating tails by the same power of b, so it rotates both length-L words together and leaves their position-wise match proportion unchanged.
Remark 3 (Common division depth). We compare every fraction at the same long-division depth. We do not restart each fraction when its own shortest prefix ends. That phase reset defines a different statistic. The common-depth convention reads every floor evaluation at the same position. For example, in base 2, \frac16=0.0\overline{01}_2,\qquad \frac13=0.\overline{01}_2,\qquad \frac23=0.\overline{10}_2. Resetting the repeating blocks makes 1/6 match 1/3. Reading all three fractions after one common digit makes 1/6 match 2/3 instead.
One step of long division turns a remainder into a digit and a new remainder. The next function records the digit.
Definition 4 (Digit function). For a prime p not dividing b, the digit function is \delta(r) \;=\; \left\lfloor \frac{br}{p} \right\rfloor, \qquad r \in \{0, 1, \ldots, p-1\}. The floor function partitions \{1, \ldots, p-1\} into contiguous bins of nearly equal size [6].
Lemma 5 (Digit-function injectivity). If p \le b + 1, then \delta is injective on \{1, \ldots, p-1\}.
Proof. Two cases.
Case p = b + 1. For r \in \{1, \ldots, b\}, \delta(r) \;=\; \left\lfloor \frac{br}{b+1} \right\rfloor \;=\; \left\lfloor r - \frac{r}{b+1} \right\rfloor \;=\; r - 1, since 0 < r/(b+1) < 1 for 1 \le r \le b. The map r \mapsto r - 1 is a bijection from \{1, \ldots, b\} to \{0, \ldots, b-1\}.
Case p \le b. For distinct r, s \in \{1, \ldots, p-1\} with r<s, we have s-r\ge1. Coprimality rules out b=p, so p\le b actually gives b>p. Therefore \frac{bs}{p}-\frac{br}{p} =\frac{b(s-r)}p \ge\frac bp>1. The two quantities before flooring are separated by more than one, so their floors are strictly ordered. ◻
Remark 6. When p > b + 1, the function \delta maps p - 1 > b distinct remainders into b digit values. By the pigeonhole principle [7], \delta cannot be injective.
Definition 7 (Digit-partitioning primes). A prime p is digit-partitioning in base b if, for every pair of distinct numerators k,\ell\in\{1,\ldots,p-1\}, the base-b expansions of k/p and \ell/p have different digits at every common position. Positions are phase-aligned from the radix point. A nontrivial cyclic rotation is therefore a different repetend phase, not an identical repetend.
Theorem 8 (Digit-partitioning characterization). A prime p not dividing b is digit-partitioning in base b if and only if p \le b + 1.
Proof. At position j\in\{0,\ldots,L-1\} of the repetend of k/p, the digit is \delta(b^{j} k \bmod p). For distinct fractions k/p and \ell/p with k \not\equiv \ell \pmod p, the arguments b^{j} k and b^{j} \ell are distinct modulo p, since b^{j} is invertible.
If p \le b + 1, Lemma 5 makes \delta injective on \{1, \ldots, p-1\}, so distinct arguments produce distinct digits at every position. Hence p is digit-partitioning.
If p > b + 1, then \delta maps p - 1 > b remainders into b values, and the pigeonhole principle yields distinct r, s \in \{1, \ldots, p-1\} with \delta(r) = \delta(s). The first repetend digits of r/p and s/p agree. Thus p is not digit-partitioning. ◻
Proposition 9 (Multiplicative cycle partition). Let n = pm with m a b-supported integer and p \nmid b. The non-terminating fractions in \{k/n : 1 \le k \le n-1\} partition into (p-1)/L families of mL fractions each, where L = \mathop{\mathrm{ord}}_p(b). Every synchronized repetend has length L. Within each family the repetends are cyclic shifts of one another; repetends from different families are not cyclic shifts.
Proof. The nonzero multiples of p are p,2p,\ldots,(m-1)p. Their reduced denominators divide m, so they terminate. If p\nmid k, the reduced denominator still has the factor p and the expansion does not terminate. Thus there are m-1 terminating fractions and (p-1)m nonterminating ones.
Choose t with m \mid b^{t} and write b^{t} = mu, where \gcd(u, p) = 1 since m is b-supported and p \nmid b. Then \frac{k}{pm} \;=\; \frac{ku}{p \cdot b^{t}}. After the common clearing depth t, the periodic tail is generated by the nonzero residue ku\bmod p. Multiplication by u permutes the nonzero residues. Each nonzero residue class modulo p has exactly m representatives in \{1,\ldots,pm-1\}. Thus each fixed starting phase occurs exactly m times; a full orbit of L phases occurs mL times.
Starting from any nonzero residue r, the remainder returns to r after q digits exactly when b^qr\equiv r\pmod p. Since r is invertible, this is equivalent to b^q\equiv1\pmod p. A repeating block of length q represents a rational whose reduced denominator divides b^q-1. Since the tail has reduced denominator p, a shorter block would force p\mid b^q-1, and hence an earlier return. Every nonterminating tail therefore has least period L=\mathop{\mathrm{ord}}_p(b).
The residues 1, \ldots, p-1 partition into (p-1)/L cosets of \langle b \rangle in (\mathbb{Z}/p\mathbb{Z})^{*}. Residues in the same coset produce cyclic shifts of a single repetend; residues in different cosets produce distinct repetends, since their orbits under multiplication by b are disjoint. A coset has L residues, and every residue occurs m times. Hence each family has mL fractions. ◻
Theorem 10 (Exact alignment formula). If p\nmid b is prime, p \le b + 1, and m \ge 1 is b-supported, then \alpha_b(pm) \;=\; \frac{2m - 1}{pm - 1}.
Proof. Choose t with m\mid b^t and put u=b^t/m. At synchronized position j\in\{0,\ldots,L-1\}, the reference tail has remainder r_j\equiv b^ju\pmod p, while the tail of k/(pm) has remainder kr_j\pmod p. The remainder r_j is nonzero. Therefore the two remainders agree if and only if k\equiv1\pmod p.
If k\equiv1\pmod p, every synchronized digit agrees with the reference. If k\not\equiv0,1\pmod p, the two remainders are distinct at every synchronized position. Lemma 5 makes the digit function injective, so none of their digits agree, whether the two phases lie in different cosets or in the same one.
The count now splits into three cases.
The m - 1 terminating fractions (multiples of p). Each contributes alignment 1.
The m numerators 1,1+p,\ldots,1+(m-1)p, all congruent to 1 modulo p. Each contributes alignment 1.
The remaining (p-2)m non-terminating fractions. Each contributes alignment 0.
Thus exactly (m-1)+m=2m-1 of the pm-1 fractions are aligned. ◻
Corollary 11 (Alignment limit). For fixed digit-partitioning p, the alignment \alpha_b(pm) \to 2/p as m \to \infty through b-supported integers.
The digit-partitioning primes in base b are precisely the primes p \le b + 1 that do not divide b.
| Base b | Digit-part. primes | Alignment limits 2/p |
|---|---|---|
| 4 | 3, 5 | 0.667,\; 0.400 |
| 6 | 5, 7 | 0.400,\; 0.286 |
| 8 | 3, 5, 7 | 0.667,\; 0.400,\; 0.286 |
| 10 | 3, 7, 11 | 0.667,\; 0.286,\; 0.182 |
| 12 | 5, 7, 11, 13 | 0.400,\; 0.286,\; 0.182,\; 0.154 |
| 16 | 3, 5, 7, 11, 13, 17 | 0.667,\; \ldots,\; 0.118 |
The class of digit-partitioning primes depends on b. The alignment formula (2m-1)/(pm-1) is uniform. It makes no reference to the base, to the order L, or to any of the mechanism-specific structure described below.
In base 10, the three digit-partitioning primes arise in structurally distinct ways. For p = 3, \mathop{\mathrm{ord}}_3(10) = 1, so repetends are single digits. For p = 7, \mathop{\mathrm{ord}}_7(10) = 6 = p-1, so 10 is a primitive root and the repetend is a cyclic number with all distinct digits. For p = 11, \mathop{\mathrm{ord}}_{11}(10) = 2 with 10 \equiv -1 \pmod{11}, so repetends form nines-complement pairs.
These three mechanisms are unified by the single condition p \le b+1 and the injectivity of \delta.
Write \varphi=(1+\sqrt{5})/2 for the golden ratio. Corollary 11 gives the alignment limit \alpha_b(pm) \to 2/p for every digit-partitioning prime. The golden threshold 1/\varphi therefore selects, within this class, precisely those primes for which 2/p > 1/\varphi, i.e. p < 2\varphi \approx 3.236. Among primes p \ge 3, only p = 3 satisfies this. It belongs to the class whenever 3\nmid b; if 3\mid b, no admissible prime p\ge3 crosses the threshold. The prime 2 is the familiar degenerate exception when the base is odd.
Why the Golden Ratio Selects the Prime Three [5] studies the golden threshold in the one-digit regime. The digit boundary places that regime inside a larger collision-free class. The same boundary p\le b+1 is present in every base.
At p=b+1, every possible digit is used and still identifies one nonzero remainder. This is the last prime size at which the digit map can do so. Since b\equiv-1\pmod p, the remainder cycles are complement pairs. In decimal, the five pairs at 11 occupy all ten digit values without sharing a position.
The same count applies to the six-cycle at 7 and the fixed repetends at 3. A matching digit identifies a matching remainder, and division carries that equality through the whole period. Different states cannot acquire an isolated match. Cycle shape is visible in the digits, but the alignment can be counted without reconstructing it.
At 13, the distinction is visible in two fractions, \begin{aligned} \frac{1}{13}&=0.\overline{076923},\\ \frac{11}{13}&=0.\overline{846153}. \end{aligned} Their digits agree in the third and sixth positions, giving score 1/3, although their remainder states remain different throughout. A shared digit no longer forces a shared trajectory. The exact formula (2m-1)/(pm-1) has reached its natural boundary because it counts complete matches and cannot include this partial agreement.
The threshold p\le b+1 specifies how much of the remainder dynamics a single digit reveals. Within it, that digit distinguishes all nonzero states. Beyond it, agreement can be temporary, and the repeating block carries information that no single digit can supply.
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