Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · 2509.04487

A symmetry approach to number tricks

Håkon Kolderup
View original source ↗Read source document ↗Export metadata ↓

Abstract

We generalize the classical "1089-number trick", which states that a certain combination of addition, subtraction and swapping the digits of a three-digit number will always output 1089. More precisely, we show that any pair of zero divisors $fg=0$ in the group ring ${\mathbb Z}[\Sigma_n]$ on the n-th symmetric group gives rise to a partition of the set of n-digit numbers into subsets $U_{\mathbf e}$ defined by linear inequalities, such that the zero divisors act constantly on each $U_{\mathbf e}$ and hence define a number trick.

Explore related subjects

math.GM

Keep this discovery

BibTeXRIS

Håkon Kolderup. 2025-08-31. A symmetry approach to number tricks. https://doi.org/10.4171/em/573

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

arXiv2026

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM↗
arXiv2026

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM↗