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Edited answer(2)

We need two horizontal cuts (parallel to the surface), these make thirds. We need one perpendicular cut to the surface making halves. And we need three more cuts perpendicular to the surface and also perpendicular to the halving cut. One cut makes 1/3 and 2/3, the second one 1/7 and 6/7 and the third one 1/4 and 3/4. And this is only 6 cuts indeed.

Why is this enough? Consider the fractions I mentioned. Because of the halving and thirding cuts made in two different perpendicular planes, it will be possible to either divide numerators or multiply denominators by 2, 3 or both (if needed). That enables us to make 71/7, 81/8 and 9*1/9 according to the needs. Obviously, the planes can be swapped. To the frosting/filling dilemma: it is impossible to remain fair with 6 cuts if the cake has only surface frosting, but it is possible, if the cake has 2 or 3 equal filling layers inside.

Previous answer:

I just realized that you are asking the number of cuts and not the resulting pieces and cutting is possible in any plane that intersects the cylinder.

If I do 3 cuts parallel to the surface making thirds and halves, then halve the cake perpendicular to the surface with 2 cuts (these cuts being also perpendicular to each other), and do 2 more cuts, one that is perpendicular to the surface and divides the cake to 1/3 and 2/3 parts and one also perpendicular to the surface that divides the cake to 3/7 and 4/7 in a similar way (the smaller portions should be opposite to each other) it is 7 cuts altogether.

Checking: What we need here is to have 71/7, 81/8 91/9 parts at the same time. The 8 equal pieces are easy to imagine. From the 1/3 part we will have 31/9 because of cutting parralel of the surface making thirds. The 2/3 part is halved by a cut (used already to generate the 8 equal parts), so it will be 21/3 and these also have their thirds. We have the remaning 61/9 parts. Similarly we will have 31/7 parts from the 3/7 part and 41/7 from the 4/7 part.

Whether this is the smallest number, I don't know.

Edited answer(2)

We need two horizontal cuts (parallel to the surface), these make thirds. We need one perpendicular cut to the surface making halves. And we need three more cuts perpendicular to the surface and also perpendicular to the halving cut. One cut makes 1/3 and 2/3, the second one 1/7 and 6/7 and the third one 1/4 and 3/4. And this is only 6 cuts indeed.

Why is this enough? Consider the fractions I mentioned. Because of the halving and thirding cuts made in two different perpendicular planes, it will be possible to either divide numerators or multiply denominators by 2, 3 or both (if needed). That enables us to make 71/7, 81/8 and 9*1/9 according to the needs. Obviously, the planes can be swapped. To the frosting/filling dilemma: it is impossible to remain fair with 6 cuts if the cake has only surface frosting, but it is possible, if the cake has 2 or 3 equal filling layers inside.

Here's a picture: (click to enlarge)

Previous answer:

I just realized that you are asking the number of cuts and not the resulting pieces and cutting is possible in any plane that intersects the cylinder.

If I do 3 cuts parallel to the surface making thirds and halves, then halve the cake perpendicular to the surface with 2 cuts (these cuts being also perpendicular to each other), and do 2 more cuts, one that is perpendicular to the surface and divides the cake to 1/3 and 2/3 parts and one also perpendicular to the surface that divides the cake to 3/7 and 4/7 in a similar way (the smaller portions should be opposite to each other) it is 7 cuts altogether.

Checking: What we need here is to have 71/7, 81/8 91/9 parts at the same time. The 8 equal pieces are easy to imagine. From the 1/3 part we will have 31/9 because of cutting parralel of the surface making thirds. The 2/3 part is halved by a cut (used already to generate the 8 equal parts), so it will be 21/3 and these also have their thirds. We have the remaning 61/9 parts. Similarly we will have 31/7 parts from the 3/7 part and 41/7 from the 4/7 part.

Whether this is the smallest number, I don't know.

Edited answer(2)

We need two horizontal cuts (parallel to the surface), these make thirds. We need one perpendicular cut to the surface making halves. And we need three more cuts perpendicular to the surface and also perpendicular to the halving cut. One cut makes 1/3 and 2/3, the second one 1/7 and 6/7 and the third one 1/4 and 3/4. And this is only 6 cuts indeed.

Why is this enough? Consider the fractions I mentioned. Because of the halving and thirding cuts made in two different perpendicular planes, it will be possible to either divide numerators or multiply denominators by 2, 3 or both (if needed). That enables us to make 71/7, 81/8 and 9*1/9 according to the needs.

Previous answer:

I just realized that you are asking the number of cuts and not the resulting pieces and cutting is possible in any plane that intersects the cylinder.

If I do 3 cuts parallel to the surface making thirds and halves, then halve the cake perpendicular to the surface with 2 cuts (these cuts being also perpendicular to each other), and do 2 more cuts, one that is perpendicular to the surface and divides the cake to 1/3 and 2/3 parts and one also perpendicular to the surface that divides the cake to 3/7 and 4/7 in a similar way (the smaller portions should be opposite to each other) it is 7 cuts altogether.

Checking: What we need here is to have 71/7, 81/8 91/9 parts at the same time. The 8 equal pieces are easy to imagine. From the 1/3 part we will have 31/9 because of cutting parralel of the surface making thirds. The 2/3 part is halved by a cut (used already to generate the 8 equal parts), so it will be 21/3 and these also have their thirds. We have the remaning 61/9 parts. Similarly we will have 31/7 parts from the 3/7 part and 41/7 from the 4/7 part.

Whether this is the smallest number, I don't know.

Edited answer(2)

We need two horizontal cuts (parallel to the surface), these make thirds. We need one perpendicular cut to the surface making halves. And we need three more cuts perpendicular to the surface and also perpendicular to the halving cut. One cut makes 1/3 and 2/3, the second one 1/7 and 6/7 and the third one 1/4 and 3/4. And this is only 6 cuts indeed.

Why is this enough? Consider the fractions I mentioned. Because of the halving and thirding cuts made in two different perpendicular planes, it will be possible to either divide numerators or multiply denominators by 2, 3 or both (if needed). That enables us to make 71/7, 81/8 and 9*1/9 according to the needs. Obviously, the planes can be swapped. To the frosting/filling dilemma: it is impossible to remain fair with 6 cuts if the cake has only surface frosting, but it is possible, if the cake has 2 or 3 equal filling layers inside.

Previous answer:

I just realized that you are asking the number of cuts and not the resulting pieces and cutting is possible in any plane that intersects the cylinder.

If I do 3 cuts parallel to the surface making thirds and halves, then halve the cake perpendicular to the surface with 2 cuts (these cuts being also perpendicular to each other), and do 2 more cuts, one that is perpendicular to the surface and divides the cake to 1/3 and 2/3 parts and one also perpendicular to the surface that divides the cake to 3/7 and 4/7 in a similar way (the smaller portions should be opposite to each other) it is 7 cuts altogether.

Checking: What we need here is to have 71/7, 81/8 91/9 parts at the same time. The 8 equal pieces are easy to imagine. From the 1/3 part we will have 31/9 because of cutting parralel of the surface making thirds. The 2/3 part is halved by a cut (used already to generate the 8 equal parts), so it will be 21/3 and these also have their thirds. We have the remaning 61/9 parts. Similarly we will have 31/7 parts from the 3/7 part and 41/7 from the 4/7 part.

Whether this is the smallest number, I don't know.

Edited answer(2)

We need two horizontal cuts (parallel to the surface), these make thirds. We need one perpendicular cut to the surface making halves. And we need three more cuts perpendicular to the surface and also perpendicular to the halving cut. One cut makes 1/3 and 2/3, the second one 1/7 and 6/7 and the third one 1/4 and 3/4. And this is only 6 cuts indeed.

Previous answer:

I just realized that you are asking the number of cuts and not the resulting pieces and cutting is possible in any plane that intersects the cylinder.

If I do 3 cuts parallel to the surface making thirds and halves, then halve the cake perpendicular to the surface with 2 cuts (these cuts being also perpendicular to each other), and do 2 more cuts, one that is perpendicular to the surface and divides the cake to 1/3 and 2/3 parts and one also perpendicular to the surface that divides the cake to 3/7 and 4/7 in a similar way (the smaller portions should be opposite to each other) it is 7 cuts altogether.

Checking: What we need here is to have 71/7, 81/8 91/9 parts at the same time. The 8 equal pieces are easy to imagine. From the 1/3 part we will have 31/9 because of cutting parralel of the surface making thirds. The 2/3 part is halved by a cut (used already to generate the 8 equal parts), so it will be 21/3 and these also have their thirds. We have the remaning 61/9 parts. Similarly we will have 31/7 parts from the 3/7 part and 41/7 from the 4/7 part.

Whether this is the smallest number, I don't know.

Edited answer(2)

We need two horizontal cuts (parallel to the surface), these make thirds. We need one perpendicular cut to the surface making halves. And we need three more cuts perpendicular to the surface and also perpendicular to the halving cut. One cut makes 1/3 and 2/3, the second one 1/7 and 6/7 and the third one 1/4 and 3/4. And this is only 6 cuts indeed.

Why is this enough? Consider the fractions I mentioned. Because of the halving and thirding cuts made in two different perpendicular planes, it will be possible to either divide numerators or multiply denominators by 2, 3 or both (if needed). That enables us to make 71/7, 81/8 and 9*1/9 according to the needs.

Previous answer:

I just realized that you are asking the number of cuts and not the resulting pieces and cutting is possible in any plane that intersects the cylinder.

If I do 3 cuts parallel to the surface making thirds and halves, then halve the cake perpendicular to the surface with 2 cuts (these cuts being also perpendicular to each other), and do 2 more cuts, one that is perpendicular to the surface and divides the cake to 1/3 and 2/3 parts and one also perpendicular to the surface that divides the cake to 3/7 and 4/7 in a similar way (the smaller portions should be opposite to each other) it is 7 cuts altogether.

Checking: What we need here is to have 71/7, 81/8 91/9 parts at the same time. The 8 equal pieces are easy to imagine. From the 1/3 part we will have 31/9 because of cutting parralel of the surface making thirds. The 2/3 part is halved by a cut (used already to generate the 8 equal parts), so it will be 21/3 and these also have their thirds. We have the remaning 61/9 parts. Similarly we will have 31/7 parts from the 3/7 part and 41/7 from the 4/7 part.

Whether this is the smallest number, I don't know.