Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 3 / c / iii / Solution

An $e^k$-runner abacus separates bead positions by their residue modulo $e^k$. Write such a residue in base $e$ asTaking one $e$-quotient sorts beads by $a_0$ and divides their positions by $e$; applying the operation again sorts by $a_1$, and so on. After $k$ stages, the iterated construction has selected exactly the same $e^k$ residue classes as the single $e^k$-quotient. The two conventional orderings may list the base-$e$ digits in opposite order, producing only a permutation of components.

Equivalently, induction on $k$ applies the same argument to every component of $Q_e(\lambda )$ and identifies the resulting $e^{k+1}$ runner partitions. Hence the iterated quotient equals a power quotient up to permutation statement is