Divisors of All Reversed Multiples

Abstract

Divisors of B squared minus one are exactly the positive divisors of all their reversed base-B multiples.

Definition 1.1 (Reversal in a base).

$$\forall B\in \mathbb{N},\forall m\in \mathbb{N},\mathrm{reverseBase}\left(B,m\right)=\mathrm{ofDigits}\left(B,\mathrm{reverse}\left(\mathrm{digits}\left(B,m\right)\right)\right)$$

Formalization. D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.reverseBase (✓ std3).

Citation. Mohamed Ayad; Rachid Bouchenna (2025). Which Integer Divides the Reverse of Any of Its Multiples?. DOI: 10.5281/zenodo.15283699. URL: https://math.colgate.edu/~integers/z37/z37.pdf.

Commentary.

Ayad and Bouchenna, section 4.1, p. 8: “For every positive integer $m=a_k{B}^k+\cdot \cdot \cdot +a_{1}B+a_0$, we call the integer $m_B^{\ast }=a_0{B}^k+a_1{B}^{k-1}+\cdot \cdot \cdot +a_k$. the reverse of $m$ in base $B$”

Here digits(B,m) is Nat.digits B m, in least-significant-first order; reverse is List.reverse and ofDigits is Nat.ofDigits. Reversing and evaluating therefore gives the displayed source expression. Trailing zeros of m become leading zeros of its reversal and contribute zero. The definition is total on natural B and m; the theorem restricts B to at least two and m to positive multiples.

Definition 1.2 (Divisibility of every reversed multiple).

$$\forall B\in \mathbb{N},\forall n\in \mathbb{N},\mathrm{HasReverseMultipleProperty}\left(B,n\right)=(\forall m\in \mathbb{N},(0<m)\Rightarrow ((n\mid m)\Rightarrow (n\mid \mathrm{reverseBase}\left(B,m\right))))$$

Formalization. D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.HasReverseMultipleProperty (✓ std3).

Citation. Mohamed Ayad; Rachid Bouchenna (2025). Which Integer Divides the Reverse of Any of Its Multiples?. DOI: 10.5281/zenodo.15283699. URL: https://math.colgate.edu/~integers/z37/z37.pdf.

Commentary.

Ayad and Bouchenna, section 4.1, p. 8: “A positive integer $n$ is said to have the property $P_B^{\ast }$ if $n$ divides $m_B^{\ast }$ for any positive multiple $m$ of $n$, that is, if for every positive integer $m$, if $n$ divides $m$, then $n$ divides $m_B^{\ast }$.”

The predicate quantifies over all natural m with 0 < m and n dividing m. Its parameters B and n are natural numbers; positivity of n and B ≥ 2 are explicit hypotheses of the characterization.

Theorem 1.3 (The characterization in every base).

$$\forall B\in \mathbb{N},(2\le B)\Rightarrow (\forall n\in \mathbb{N},(0<n)\Rightarrow (\mathrm{HasReverseMultipleProperty}\left(B,n\right)\iff n\mid B^2-1))$$

Proof. Machine-checked in Lean as D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.result (✓ std3). ∎

Resolves. Problems/ayad-bouchenna-base-b-reverse-multiple-divisors (proved) by D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.result.

Source. Repository-derived.

Acknowledgement. Mohamed Ayad; Rachid Bouchenna (2025). Which Integer Divides the Reverse of Any of Its Multiples?. DOI: 10.5281/zenodo.15283699. URL: https://math.colgate.edu/~integers/z37/z37.pdf.

Commentary.

Ayad and Bouchenna, Problem 1, p. 9: “Let $B\ge 2$ be any number base. Are the divisors of $B^2-1$ the only positive integers satisfying the property $P_B^{\ast }$ ?”

The answer is yes for every natural base B ≥ 2 and every positive natural n. The implication from divisibility by B² − 1 is Proposition 5 on p. 8. The converse follows by first forcing n to be coprime to B using a multiple with leading digit one. If B² is not one modulo n, the multiplicative order T of B is at least two. Sparse digit lists have ones at positions 0, 1, T, …, (c+1)T. Their forward and reversed residues force B² = 1 modulo n. The vertical bars in the statement denote divisibility; B² − 1 is natural subtraction, agreeing with integer subtraction under B ≥ 2.

References

• Truth anchor: D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.HasReverseMultipleProperty
• Truth anchor: D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.result
• Truth anchor: D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.reverseBase