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My second answer is

6 cuts if the peices can be moved before each cut.

Suppose the cake is first cut into 9 equal pieces.
If only 8 guests arrive, one spare piece must also be cut into 8 equal portions.
If only 7 guests arrive, two spare pieces must also be cut into 7 portions each.

The first 3 cuts are (not always through the centre):
1 $\frac{4}{9} \frac{5}{9} $
2 $\frac{2}{9} \frac{2}{9} \frac{2}{9} \frac{3}{9} $
3 $\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{2}{9} $

Put 5 of the pieces to one side, and work with the 4 remaining pieces. The largest needs to be cut into 2 portions, another into 8 portions and the other two into 7 portions.
After cut 4 you can put the two $\frac{1}{9}$ pieces to one side, leaving (as fractions of those parts):
4 $ \frac{4}{8} \frac{4}{8} \frac{4}{7} \frac{3}{7} \frac{4}{7} \frac{3}{7} $
5 $ \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} $

Put the two $\frac{1}{7} $ pieces to one side, and halve the remaining ones:
6 $ \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7}$

This produces:
6 pieces of size $ \frac{1}{9}$
8 pieces of size $ \frac{1}{72}$
14 pieces of size $ \frac{1}{63}$
28 pieces in total with $ \frac{6}{9} + \frac{8}{72} + \frac{14}{63} = 1 $

If 9 guests arrive, 6 get $(\frac{1}{9})$, 2 each $( \frac{7}{63} = \frac{1}{9})$, and 1 gets $( \frac{8}{72} = \frac{1}{9})$ (28 pieces).

If 8 guests arrive, 6 get $(\frac{1}{9} + \frac{1}{72} = \frac{1}{8})$, 2 get $(\frac{7}{63} + \frac{1}{72} = \frac{1}{8})$ (28 pieces).

If 7 guests arrive, 6 get $(\frac{1}{9} + \frac{2}{63} = \frac{1}{7})$, 1 gets $( \frac{2}{63} + \frac{8}{72} = \frac{1}{7})$ (28 pieces).

But how is the cake to be divided in the ratio 4/5 with the first cut? It's possible, but other every cut will be an awkward calculation too. But with 1 extra cut and a protractor, every piece can be a regular shaped slice. If the first division is made from 2 half cuts, at an angle of 160° then every piece can be a true sector shape. In that case it needs
7 cuts.

---

My first answer was >! **9 cuts** >! >! The lowest common multiple of the possible guests is $7 \times 8 \times 9 = 504$. >! If you move pieces so that every cut halves every piece, **9 cuts** will make $2^9 = 512$ pieces. >! >! If $8$ guests arrive, you serve $8 \times 64 = 512$ pieces. >! >! The others are more tricky. Based on the assumption that cake is soft and crumbly, and it is not possible to cut pieces exactly in half, then: >! >! For $7$ guests, if you serve $7 \times 73 = 511$ pieces, there would be one piece remaining. So serve the $74$ smallest pieces to one guest, and $73$ at random to each of the other guests. >! >! For $9$ guests, if you serve $9 \times 57 = 513$ pieces, there would be one piece short. So serve the $56$ largest pieces to one guest, and $57$ at random to each of the other guests. >! >! To make it easy when the guests arrive, you can already have sorted out the $74$ smallest pieces into one pile and the $56$ largest pieces into another pile. Depending on how many guests arrive, you dump one or both of these piles into the main heap.

My second answer is

6 cuts if the peices can be moved before each cut.

Suppose the cake is first cut into 9 equal pieces.
If only 8 guests arrive, one spare piece must also be cut into 8 equal portions.
If only 7 guests arrive, two spare pieces must also be cut into 7 portions each.

The first 3 cuts are (not always through the centre):
1 $\frac{4}{9} \frac{5}{9} $
2 $\frac{2}{9} \frac{2}{9} \frac{2}{9} \frac{3}{9} $
3 $\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{2}{9} $

Put 4 of the pieces to one side, and work with the 4 remaining pieces. The largest needs to be cut into 2 portions, another into 8 portions and the other two into 7 portions.
After cut 4 you can put the two $\frac{1}{9}$ pieces to one side, leaving (as fractions of those parts):
4 $ \frac{4}{8} \frac{4}{8} \frac{4}{7} \frac{3}{7} \frac{4}{7} \frac{3}{7} $
5 $ \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} $

Put the two $\frac{1}{7} $ pieces to one side, and halve the remaining ones:
6 $ \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7}$

This produces:
6 pieces of size $ \frac{1}{9}$
8 pieces of size $ \frac{1}{72}$
14 pieces of size $ \frac{1}{63}$
28 pieces in total with $ \frac{6}{9} + \frac{8}{72} + \frac{14}{63} = 1 $

If 9 guests arrive, 6 get $(\frac{1}{9})$, 2 each $( \frac{7}{63} = \frac{1}{9})$, and 1 gets $( \frac{8}{72} = \frac{1}{9})$ (28 pieces).

If 8 guests arrive, 6 get $(\frac{1}{9} + \frac{1}{72} = \frac{1}{8})$, 2 get $(\frac{7}{63} + \frac{1}{72} = \frac{1}{8})$ (28 pieces).

If 7 guests arrive, 6 get $(\frac{1}{9} + \frac{2}{63} = \frac{1}{7})$, 1 gets $( \frac{2}{63} + \frac{8}{72} = \frac{1}{7})$ (28 pieces).

But how is the cake to be divided in the ratio 4/5 with the first cut? It's possible, but other every cut will be an awkward calculation too. But with 1 extra cut and a protractor, every piece can be a regular shaped slice. If the first division is made from 2 half cuts, at an angle of 160° then every piece can be a true sector shape. In that case it needs
7 cuts.

---

My first answer was >! **9 cuts** >! >! The lowest common multiple of the possible guests is $7 \times 8 \times 9 = 504$. >! If you move pieces so that every cut halves every piece, **9 cuts** will make $2^9 = 512$ pieces. >! >! If $8$ guests arrive, you serve $8 \times 64 = 512$ pieces. >! >! The others are more tricky. Based on the assumption that cake is soft and crumbly, and it is not possible to cut pieces exactly in half, then: >! >! For $7$ guests, if you serve $7 \times 73 = 511$ pieces, there would be one piece remaining. So serve the $74$ smallest pieces to one guest, and $73$ at random to each of the other guests. >! >! For $9$ guests, if you serve $9 \times 57 = 513$ pieces, there would be one piece short. So serve the $56$ largest pieces to one guest, and $57$ at random to each of the other guests. >! >! To make it easy when the guests arrive, you can already have sorted out the $74$ smallest pieces into one pile and the $56$ largest pieces into another pile. Depending on how many guests arrive, you dump one or both of these piles into the main heap.

My second answer is

6 cuts if the peices can be moved before each cut.

Suppose the cake is first cut into 9 equal pieces.
If only 8 guests arrive, one spare piece must also be cut into 8 equal portions.
If only 7 guests arrive, two spare pieces must also be cut into 7 portions each.

The first 3 cuts are (not always through the centre):
1 $\frac{4}{9} \frac{5}{9} $
2 $\frac{2}{9} \frac{2}{9} \frac{2}{9} \frac{3}{9} $
3 $\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{2}{9} $

Put 5 of the pieces to one side, and work with the 4 remaining pieces. The largest needs to be cut into 2 portions, another into 8 portions and the other two into 7 portions.
After cut 4 you can put the two $\frac{1}{9}$ pieces to one side, leaving (as fractions of those parts):
4 $ \frac{4}{8} \frac{4}{8} \frac{4}{7} \frac{3}{7} \frac{4}{7} \frac{3}{7} $
5 $ \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} $

Put the two $\frac{1}{7} $ pieces to one side, and halve the remaining ones:
6 $ \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7}$

This produces:
6 pieces of size $ \frac{1}{9}$
8 pieces of size $ \frac{1}{72}$
14 pieces of size $ \frac{1}{63}$
28 pieces in total with $ \frac{6}{9} + \frac{8}{72} + \frac{14}{63} = 1 $

If 9 guests arrive, 6 get $(\frac{1}{9})$, 2 each $( \frac{7}{63} = \frac{1}{9})$, and 1 gets $( \frac{8}{72} = \frac{1}{9})$ (28 pieces).

If 8 guests arrive, 6 get $(\frac{1}{9} + \frac{1}{72} = \frac{1}{8})$, 2 get $(\frac{7}{63} + \frac{1}{72} = \frac{1}{8})$ (28 pieces).

If 7 guests arrive, 6 get $(\frac{1}{9} + \frac{2}{63} = \frac{1}{7})$, 1 gets $( \frac{2}{63} + \frac{8}{72} = \frac{1}{7})$ (28 pieces).

---

My first answer was >! **9 cuts** >! >! The lowest common multiple of the possible guests is $7 \times 8 \times 9 = 504$. >! If you move pieces so that every cut halves every piece, **9 cuts** will make $2^9 = 512$ pieces. >! >! If $8$ guests arrive, you serve $8 \times 64 = 512$ pieces. >! >! The others are more tricky. Based on the assumption that cake is soft and crumbly, and it is not possible to cut pieces exactly in half, then: >! >! For $7$ guests, if you serve $7 \times 73 = 511$ pieces, there would be one piece remaining. So serve the $74$ smallest pieces to one guest, and $73$ at random to each of the other guests. >! >! For $9$ guests, if you serve $9 \times 57 = 513$ pieces, there would be one piece short. So serve the $56$ largest pieces to one guest, and $57$ at random to each of the other guests. >! >! To make it easy when the guests arrive, you can already have sorted out the $74$ smallest pieces into one pile and the $56$ largest pieces into another pile. Depending on how many guests arrive, you dump one or both of these piles into the main heap.

My second answer is

6 cuts if the peices can be moved before each cut.

Suppose the cake is first cut into 9 equal pieces.
If only 8 guests arrive, one spare piece must also be cut into 8 equal portions.
If only 7 guests arrive, two spare pieces must also be cut into 7 portions each.

The first 3 cuts are (not always through the centre):
1 $\frac{4}{9} \frac{5}{9} $
2 $\frac{2}{9} \frac{2}{9} \frac{2}{9} \frac{3}{9} $
3 $\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{2}{9} $

Put 5 of the pieces to one side, and work with the 4 remaining pieces. The largest needs to be cut into 2 portions, another into 8 portions and the other two into 7 portions.
After cut 4 you can put the two $\frac{1}{9}$ pieces to one side, leaving (as fractions of those parts):
4 $ \frac{4}{8} \frac{4}{8} \frac{4}{7} \frac{3}{7} \frac{4}{7} \frac{3}{7} $
5 $ \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} $

Put the two $\frac{1}{7} $ pieces to one side, and halve the remaining ones:
6 $ \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7}$

This produces:
6 pieces of size $ \frac{1}{9}$
8 pieces of size $ \frac{1}{72}$
14 pieces of size $ \frac{1}{63}$
28 pieces in total with $ \frac{6}{9} + \frac{8}{72} + \frac{14}{63} = 1 $

If 9 guests arrive, 6 get $(\frac{1}{9})$, 2 each $( \frac{7}{63} = \frac{1}{9})$, and 1 gets $( \frac{8}{72} = \frac{1}{9})$ (28 pieces).

If 8 guests arrive, 6 get $(\frac{1}{9} + \frac{1}{72} = \frac{1}{8})$, 2 get $(\frac{7}{63} + \frac{1}{72} = \frac{1}{8})$ (28 pieces).

If 7 guests arrive, 6 get $(\frac{1}{9} + \frac{2}{63} = \frac{1}{7})$, 1 gets $( \frac{2}{63} + \frac{8}{72} = \frac{1}{7})$ (28 pieces).

But how is the cake to be divided in the ratio 4/5 with the first cut? It's possible, but other every cut will be an awkward calculation too. But with 1 extra cut and a protractor, every piece can be a regular shaped slice. If the first division is made from 2 half cuts, at an angle of 160° then every piece can be a true sector shape. In that case it needs
7 cuts.

---

My first answer was >! **9 cuts** >! >! The lowest common multiple of the possible guests is $7 \times 8 \times 9 = 504$. >! If you move pieces so that every cut halves every piece, **9 cuts** will make $2^9 = 512$ pieces. >! >! If $8$ guests arrive, you serve $8 \times 64 = 512$ pieces. >! >! The others are more tricky. Based on the assumption that cake is soft and crumbly, and it is not possible to cut pieces exactly in half, then: >! >! For $7$ guests, if you serve $7 \times 73 = 511$ pieces, there would be one piece remaining. So serve the $74$ smallest pieces to one guest, and $73$ at random to each of the other guests. >! >! For $9$ guests, if you serve $9 \times 57 = 513$ pieces, there would be one piece short. So serve the $56$ largest pieces to one guest, and $57$ at random to each of the other guests. >! >! To make it easy when the guests arrive, you can already have sorted out the $74$ smallest pieces into one pile and the $56$ largest pieces into another pile. Depending on how many guests arrive, you dump one or both of these piles into the main heap.

My second answer is

6 cuts if the peices can be moved before each cut.

Suppose the cake is first cut into 9 equal pieces.
If only 8 guests arrive, one spare piece must also be cut into 8 equal portions.
If only 7 guests arrive, two spare pieces must also be cut into 7 portions each.

The first 3 cuts are (not always through the centre):
1 $\frac{4}{9} \frac{5}{9} $
2 $\frac{2}{9} \frac{2}{9} \frac{2}{9} \frac{3}{9} $
3 $\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{2}{9} $

Put 5 of the pieces to one side, and work with the 4 remaining pieces. The largest needs to be cut into 2 portions, another into 8 portions and the other two into 7 portions.
After cut 4 you can put the two $\frac{1}{9}$ pieces to one side, leaving (as fractions of those parts):
4 $ \frac{4}{8} \frac{4}{8} \frac{4}{7} \frac{3}{7} \frac{4}{7} \frac{3}{7} $
5 $ \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} $

Put the two $\frac{1}{7} $ pieces to one side, and halve the remaining ones:
6 $ \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7}$

---

My first answer was >! **9 cuts** >! >! The lowest common multiple of the possible guests is $7 \times 8 \times 9 = 504$. >! If you move pieces so that every cut halves every piece, **9 cuts** will make $2^9 = 512$ pieces. >! >! If $8$ guests arrive, you serve $8 \times 64 = 512$ pieces. >! >! The others are more tricky. Based on the assumption that cake is soft and crumbly, and it is not possible to cut pieces exactly in half, then: >! >! For $7$ guests, if you serve $7 \times 73 = 511$ pieces, there would be one piece remaining. So serve the $74$ smallest pieces to one guest, and $73$ at random to each of the other guests. >! >! For $9$ guests, if you serve $9 \times 57 = 513$ pieces, there would be one piece short. So serve the $56$ largest pieces to one guest, and $57$ at random to each of the other guests. >! >! To make it easy when the guests arrive, you can already have sorted out the $74$ smallest pieces into one pile and the $56$ largest pieces into another pile. Depending on how many guests arrive, you dump one or both of these piles into the main heap.

My second answer is

6 cuts if the peices can be moved before each cut.

Suppose the cake is first cut into 9 equal pieces.
If only 8 guests arrive, one spare piece must also be cut into 8 equal portions.
If only 7 guests arrive, two spare pieces must also be cut into 7 portions each.

The first 3 cuts are (not always through the centre):
1 $\frac{4}{9} \frac{5}{9} $
2 $\frac{2}{9} \frac{2}{9} \frac{2}{9} \frac{3}{9} $
3 $\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{2}{9} $

Put 5 of the pieces to one side, and work with the 4 remaining pieces. The largest needs to be cut into 2 portions, another into 8 portions and the other two into 7 portions.
After cut 4 you can put the two $\frac{1}{9}$ pieces to one side, leaving (as fractions of those parts):
4 $ \frac{4}{8} \frac{4}{8} \frac{4}{7} \frac{3}{7} \frac{4}{7} \frac{3}{7} $
5 $ \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{8} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} \frac{2}{7} \frac{2}{7} \frac{2}{7} \frac{1}{7} $

Put the two $\frac{1}{7} $ pieces to one side, and halve the remaining ones:
6 $ \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{8} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7} \frac{1}{7}$

This produces:
6 pieces of size $ \frac{1}{9}$
8 pieces of size $ \frac{1}{72}$
14 pieces of size $ \frac{1}{63}$
28 pieces in total with $ \frac{6}{9} + \frac{8}{72} + \frac{14}{63} = 1 $

If 9 guests arrive, 6 get $(\frac{1}{9})$, 2 each $( \frac{7}{63} = \frac{1}{9})$, and 1 gets $( \frac{8}{72} = \frac{1}{9})$ (28 pieces).

If 8 guests arrive, 6 get $(\frac{1}{9} + \frac{1}{72} = \frac{1}{8})$, 2 get $(\frac{7}{63} + \frac{1}{72} = \frac{1}{8})$ (28 pieces).

If 7 guests arrive, 6 get $(\frac{1}{9} + \frac{2}{63} = \frac{1}{7})$, 1 gets $( \frac{2}{63} + \frac{8}{72} = \frac{1}{7})$ (28 pieces).

---

My first answer was >! **9 cuts** >! >! The lowest common multiple of the possible guests is $7 \times 8 \times 9 = 504$. >! If you move pieces so that every cut halves every piece, **9 cuts** will make $2^9 = 512$ pieces. >! >! If $8$ guests arrive, you serve $8 \times 64 = 512$ pieces. >! >! The others are more tricky. Based on the assumption that cake is soft and crumbly, and it is not possible to cut pieces exactly in half, then: >! >! For $7$ guests, if you serve $7 \times 73 = 511$ pieces, there would be one piece remaining. So serve the $74$ smallest pieces to one guest, and $73$ at random to each of the other guests. >! >! For $9$ guests, if you serve $9 \times 57 = 513$ pieces, there would be one piece short. So serve the $56$ largest pieces to one guest, and $57$ at random to each of the other guests. >! >! To make it easy when the guests arrive, you can already have sorted out the $74$ smallest pieces into one pile and the $56$ largest pieces into another pile. Depending on how many guests arrive, you dump one or both of these piles into the main heap.