Removing an

e

-hook removes one cell of every residue modulo

e

. If

λ

and

μ

have the same

e

-core and the same size, part a(i) shows that they have the same

e

-weight. Their residue multisets are therefore obtained from the common core by adjoining the same number of complete residue sets

{0, 1, \dots, e - 1}

. Thus their

e

-contents are equal.

ii

Solution

The beta number in row

i

h_{i} = λ_{i} + m - i .

Since

e ∣ m

r_{e} (i, λ_{i}) \equiv λ_{i} - i \equiv h_{i} - m \equiv h_{i} (mod e) .

Pad both partitions to the same number

m

of rows, with

e ∣ m

s

, subtracting the number of cells of residue

s - 1

s

telescopes row by row. Since the row-start residues

1 - i

are the same for both padded diagrams, equality of

e

-contents and part ii imply that their beta sets contain the same number of elements on each abacus runner.

Sliding beads upward to obtain the

e

-core preserves the number of beads on every runner. A packed runner is uniquely determined by that number, so the two packed abaci coincide. Hence

C_{e} (λ) = C_{e} (μ)

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