Research Notes

Decimal endings on both faces of the grid

Yagel Bar · 2026-10-20 · Proved congruence classification

Question and answer

When do the two faces end in 09 and 11, and which other endings are possible? From row 4 onward, every cell has exactly one of four two-digit pairs: 09/11, 29/31, 69/71 or 89/91. The pair follows an exact exponent cycle of length 1, 2 or 4. Row 5 always gives 09/11; that pattern does not persist across all later rows. The first three rows have separate cycles or initial exceptions.

This is a congruence theorem for all positive orders and exponents, on both faces, regardless of primality. In particular, a decimal ending is not a primality test.

Definitions and the last digit

Let $p=p_k$ be the $k$-th prime, $h=p_{k-1}\mathrm{\#}$ with $p_0\mathrm{\#}=1$, and $M=h{p}^n$, for $k,n\ge 1$. Write $N_{-}=M-1=Y_k(n)$ and $N_{+}=M+1=Y_k^{+}(n)$. An ending is a residue modulo ${10}^d$, padded on the left with zeros to $d$ places when needed. Pairs below list minus first, plus second.

Last-digit theorem. For every $k\ge 3$ and $n\ge 1$, the endings are 9/1. The product $M$ contains both 2 and 5: at $k=3$ they come from $h=6$ and $p=5$; at $k\ge 4$ both are already in $h$. Thus $M\equiv 0(\mathrm{mod}10)$. For $k=1$, the last-digit pairs cycle as 1/3, 3/5, 7/9, 5/7; for $k=2$ they cycle as 5/7, 7/9, 3/5, 1/3, starting at $n=1$. These follow by multiplying successively by 2 or by 3 modulo 10.

Complete two-digit classification from row 4

For $k\ge 4$, set $a\equiv h/10(\mathrm{mod}10)$ and $b\equiv p(\mathrm{mod}10)$. The primorial is squarefree and includes 2 and 5 exactly once, so both $a$ and $b$ belong to $\{1,3,7,9\}$. Cancellation of a factor of 10 gives

$$M\equiv 10r(\mathrm{mod}100),r\equiv a{b}^n(\mathrm{mod}10).$$

This proves the entire classification:

The period is exactly the multiplicative order of $b$ modulo 10: it is 1 for $b=1$, 2 for $b=9$, and 4 for $b=3$ or $b=7$. Indeed, equality after a shift $T$ is equivalent to $a{b}^n(b^T-1)\equiv 0(\mathrm{mod}10)$. Since $a{b}^n$ is a unit, this is equivalent to $b^T\equiv 1(\mathrm{mod}10)$. There is no initial transient.

In particular 09/11 occurs exactly when $a{b}^n\equiv 1(\mathrm{mod}10)$. The following table solves that condition completely. A residue 0 modulo a period means its positive multiples, since $n\ge 1$ throughout.

To check the table, list $3^n(\mathrm{mod}10)$ as 3, 9, 7, 1, and $7^n$ as 7, 9, 3, 1; the powers of 9 alternate 9, 1. Multiplication by each $a$ gives exactly the displayed solutions. Thus a row has 09/11 at every exponent if and only if $a=b=1$. These conditions depend on the row's primorial and base; there is no threshold in $k$ beyond which every row has that ending. To prove the latter assertion, suppose there were only finitely many primes other than 2 and 5 whose residue modulo 10 is not 1, and let $Q$ be their product. The integer $10Q-1$ is odd, not divisible by 5 or by any prime in $Q$, and is 9 modulo 10. Yet a product of primes all 1 modulo 10 would be 1 modulo 10, a contradiction. Such bases therefore occur arbitrarily late, and every one has an exponent period greater than 1.

Examples and the exceptional rows

For example, row 4 begins $(209,211)$ and $(1469,1471)$. Row 5 begins $(2309,2311)$; row 6 begins $(30029,30031)$, disproving a universal 09/11 rule for all $k\ge 5$. The row 3 assertion follows from $6\cdot 5=30$ and $6\cdot 5^n\equiv 50(\mathrm{mod}100)$ for $n\ge 2$: $5^2\equiv 25(\mathrm{mod}100)$ and multiplication by 5 leaves 25 unchanged.

For row 1, once $n\ge 2$ the center is 0 modulo 4; modulo 25 its powers of 2 have order 20. To verify that order, $2^{10}\equiv -1(\mathrm{mod}25)$, $2^4\equiv 16(\mathrm{mod}25)$ and $2^{20}\equiv 1(\mathrm{mod}25)$ exclude every proper divisor of 20. The Chinese remainder theorem therefore gives exact period 20 modulo 100. The $n=1$ center has residue 2 modulo 4, so it never repeats later. For row 2, canceling the coefficient 2 reduces the period question to powers of 3 modulo 50. Their order is 20: $3^{10}\equiv -1(\mathrm{mod}25)$, $3^4\equiv 6(\mathrm{mod}25)$, and $3^{20}\equiv 1(\mathrm{mod}25)$, while the modulo-2 condition is automatic. None of rows 1–3 ever has 09/11: rows 1 and 2 lack a factor 5 in $M$, and row 3 has precisely the two cases just proved.

Any number of decimal places

Period theorem. Fix $d\ge 1$ decimal places. If $p$ is coprime to 10, put $g=gcd(h,{10}^d)$ and $m={10}^d/g$. Both faces have exact period ${\mathrm{ord}}_m(p)$ from $n=1$, taking the order modulo 1 to be 1.

Proof. Equality of the center's residues at $n+T$ and $n$ is equivalent to ${10}^d\mid h{p}^n(p^T-1)$. Dividing by $g$ leaves a coefficient $(h/g)p^n$ coprime to $m$, so the condition is precisely $p^T\equiv 1(\mathrm{mod}m)$. Adding or subtracting 1 preserves this equality and its least positive period. For $k\ge 4$, $g=10$, so the exact period is ${\mathrm{ord}}_{{10}^{d-1}}(p_k)$. For $k=2$, it is ${\mathrm{ord}}_{5\cdot {10}^{d-1}}(3)$.

The two non-unit bases have a transient and must be treated separately:

Proof for the exceptions. At $k=1$, the residue modulo $2^d$ is zero exactly when $n\ge d$. The independent modulo-$5^d$ component is a unit cycle of powers of 2. At $k=3$, the residue modulo $5^d$ is zero exactly when $n\ge d$. Equality in the modulo-$2^d$ component, after canceling the factor $gcd(6,2^d)=2$, is precisely $5^T\equiv 1(\mathrm{mod}{2}^{d-1})$. For $d=1$ the modulus is 1 and the condition is automatic. The Chinese remainder theorem combines each stable zero component with its unit cycle to give the stated exact period. No periodic tail can start earlier: for $n<d$, the valuation at the base prime is $n$, and no later exponent has the same residue modulo that prime's $d$-th power. Subtracting or adding 1 again preserves equality. This proves both the start and the minimal period.

For example, row 3 has three-digit endings 029/031 at $n=1$, 149/151 at $n=2$, and 749/751 for every $n\ge 3$. The eventual fixed ending is a consequence of the theorem, not a guess from a short sample.

Verification, scope and related reading

The proof applies to every cell, including composites and the value $Y_1(1)=1$, which is neither prime nor composite. It supplies no prime counts or asymptotic prime-frequency claim. The patterns follow from the specified primorial construction and elementary congruences; no novelty claim is made for the multiplicative-order or Chinese-remainder principles.

This standalone Python example checks the displayed small rows by ordinary integer arithmetic, with no period-finding algorithm:

h = 1
for k, p in enumerate((2, 3, 5, 7, 11, 13, 17, 19), 1):
    pairs = [f"{(h*p**n-1)%100:02d}/{(h*p**n+1)%100:02d}"
             for n in range(1, 9)]
    print(k, pairs)
    h *= p

The repository's tools/test_trailing_digits.py additionally checks the classification and minimal periods against direct integer values on bounded grids. These computations check the statements' implementation; the proof above establishes the unbounded claims.

The paper's structural section explains the grid's sieve and factored neighbors. Exponent sieves addresses a different modular question: how divisors exclude exponent classes, and why their survival fractions need not multiply. To compute a complete integer, use Explore the Grid.