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

Trace the southeast boundary of the Hook of a Young diagram based at $(i,j)$. At each horizontal boundary step record the hook length $h_{i,y}$ of the cell in row $i$ above that step. At each vertical step ending beside row $x$, record $h_{i,j}-h_{x,j}$. Starting at the northeast end and moving to the southwest end, these records increase by one from $1$ to $h_{i,j}$; horizontal and vertical steps are disjoint and account for every step. Therefore the Hook-interval decomposition at a Young-diagram cell is

The supplied row-hook formula isConsequently $h\in H_i(\lambda )$ exactly when $h_i-h\ge 0$ and $h_i-h$ is not one of $h_{i+1},\dots ,h_m$. Since $h_i-h\le h_i<h_j$ for $j<i$, this is equivalent to $h_i-h\notin X$. We have proved the hook criterion in a beta set

If $ef$ is a hook length, the beta-set interpretation gives a bead at some position $b$ and a gap at $b-ef$. In the finite progressionthe first position is occupied and the last is empty. Some consecutive pair is therefore a bead followed by a gap. Their distance is $e$, so the criterion gives a hook of length $e$. This proves the divisor closure of hook lengths.

LetA direct comparison of the affected bead-gap pairs shows that removing a rim 2-hook preserves $D$: the pairs whose parities change cancel in odd-even pairs. Repeating this removal givesEvery 2-core is a staircaseAll hook lengths in this staircase are odd, and it has $1+2+\cdots +r=r(r+1)/2$ cells. Hence the odd-minus-even hook count of a partition isThus the requested integer is $m=r+1=\ell (C_2(\lambda ))+1$.

Use four beads, for which $(3,1)$ has beta setOn runners of residues $0,1,2,3$, division by four gives respectively the beta setsOnly $\{1\}$ represents a nonempty partition, namely $(1)$. Therefore the four-quotient of the partition three-one is

A two-runner partition abacus givesand the 2-quotients of $(1)$ and $\varnothing$ are empty. Thus the two-quotient tower of the partition three-one has nonempty levels

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