For $\lambda=(6,5,3)$ use the three-bead <beta set> $\{8,6,3\}$. Its runners give the 3-<quotient of a partition>
$$
Q_3(\lambda)=\bigl((1,1),\varnothing,(2)\bigr).
$$
The hook lengths divisible by three and their images are
$$
\begin{aligned}
f(H_{1,3}(\lambda))&=H_{1,1}(\lambda^{(2)}),&6&=3\cdot2,\\
f(H_{1,5}(\lambda))&=H_{1,2}(\lambda^{(2)}),&3&=3\cdot1,\\
f(H_{2,1}(\lambda))&=H_{1,1}(\lambda^{(0)}),&6&=3\cdot2,\\
f(H_{3,1}(\lambda))&=H_{2,1}(\lambda^{(0)}),&3&=3\cdot1.
\end{aligned}
$$
There are no others, as the complete hook-length rows are $(8,7,6,4,3,1)$, $(6,5,4,2,1)$, and $(3,2,1)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 3 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-160/3/a/ii/solution