Well, no. A (general) prism by definition has parallelograms as side faces. The "rectangular" in "rectangular prism" only refers to the base polygons. These are rectangles and they have right angles, of course. But there are also "oblique/slanted prisms" where the edges and faces joining the base polygons are not perpendicular to the base polygons.
Yet, come to think about it, the problem might not be well posed without the distinction of right rectangular prisms (but I'm not sure about this).
When there are non-rectangular faces, at least two of the diagonals would not be of length 2024, so if they can be arbitrarily long, there's no upper limit to the volume.
But it might be fun to find the minimum volume when the longest diagonal is 2024. With sides of 1, 1, and 2023, I get √(8091/4) as the minimum volume.
That was interesting. I suppose there exists an oblique rectangular prism with sides 1, 1, 2023, longest diagonal 2024, and arbitrarily small volume.
For arbitrarily small V, let x = (sqrt(33129022255 - 16370120 V^2) + 4045)/8185060, y = (16554279805 - sqrt(33129022255 - 16370120 V^2))/16558376380, and z = V/2023. Taking points A(0, 0, 0), B(1, 0, 0), C(1, 2023, 0), D(0, 2023, 0), E(x, y, z), F(x + 1, y, z), G(x + 1, y + 2023, z), H(x, y + 2023, z) in coordinate space, the oblique rectangular prism ABCD-EFGH has sides AB = 1, AE = 1, AD = 2023, the longest diagonal AG = 2024, and the volume V.
I see. I got my minimum by considering diagonals to the intersection of a sphere of radius 2024 and a sphere of radius 2023 displaced by √2, which is a circle, but the parts of the circle corresponding to smaller volumes also correspond to having another diagonal longer than 2024, and my minimum is the four diagonals that correspond to having two diagonals of length 2024.
But, as you point out, there are also diagonals to the intersection of a sphere of radius 2024 and a sphere of radius 1 displaced by √(2023²+1), giving arbitrarily small volumes.
With no puzzles this week, I decided to try last week's puzzle as applied to next year (2025).
1. Knowing that the rectangular prism with the largest volume for a given internal diagonal is a cube, I find that the sides of the cube should be 2025/√3 = 1169.134, the maximum volume (noninteger x,y,z) is (2025/√3)³ = 1598060439.6271. To find the closest volume to this maximum for integer x,y,z, I plotted a spreadsheet table where x steps down from 1169, y steps up from 1169, and the table is filled with the possible z values, z = √(2025² - x² - y²) and I look for integer values of z in the table. I find 2 candidates:
x * y * z = 1194 * 1158 * 1155 = 1596963060, and
x * y * z = 1183 * 1144 * 1180 = 1596955360
The first of these two is slightly bigger, so my answer to the first puzzle is:
x * y * z = 1194 * 1158 * 1155 = 1596963060 <<======================
2. 2025 has quite a long Collatz sequence, 157 numbers. A Collatz sequence that contains 2025 must therefore have more than 157 numbers. I started with numbers less than 2025 and worked down, looking for one with a Collatz sequence of more than 157 numbers. I looked only at those Collatz sequences longer than 157 and it appears to me that the smallest number with 2025 in its Collatz sequence is, ... 2025. A surprise considering the vast number of numbers in the Collatz sequences for numbers between 1 and 2025.
Taking another approach to the Collatz sequence puzzle, I considered what possible numbers could precede 2025 in a Collatz sequence. It appears that the preceding numbers must be 2025*2ⁿ for n=1,2,3...inf. Exponentially increasing numbers with never a number less than 2025. It appears that any year evenly divisible by 3 (like 2025) has the property that its Collatz sequence is not contained in the Collatz sequence of a lesser number.
I'm a little confused by the term "rectangular prism". Can I assume that this is the same as a rectangular cuboid, i.e. a right rectangular prism?
"Rectangular prism" implies a right rectangular prism, because all the rectangles must have right angles.
Well, no. A (general) prism by definition has parallelograms as side faces. The "rectangular" in "rectangular prism" only refers to the base polygons. These are rectangles and they have right angles, of course. But there are also "oblique/slanted prisms" where the edges and faces joining the base polygons are not perpendicular to the base polygons.
Yet, come to think about it, the problem might not be well posed without the distinction of right rectangular prisms (but I'm not sure about this).
When there are non-rectangular faces, at least two of the diagonals would not be of length 2024, so if they can be arbitrarily long, there's no upper limit to the volume.
But it might be fun to find the minimum volume when the longest diagonal is 2024. With sides of 1, 1, and 2023, I get √(8091/4) as the minimum volume.
That was interesting. I suppose there exists an oblique rectangular prism with sides 1, 1, 2023, longest diagonal 2024, and arbitrarily small volume.
For arbitrarily small V, let x = (sqrt(33129022255 - 16370120 V^2) + 4045)/8185060, y = (16554279805 - sqrt(33129022255 - 16370120 V^2))/16558376380, and z = V/2023. Taking points A(0, 0, 0), B(1, 0, 0), C(1, 2023, 0), D(0, 2023, 0), E(x, y, z), F(x + 1, y, z), G(x + 1, y + 2023, z), H(x, y + 2023, z) in coordinate space, the oblique rectangular prism ABCD-EFGH has sides AB = 1, AE = 1, AD = 2023, the longest diagonal AG = 2024, and the volume V.
I see. I got my minimum by considering diagonals to the intersection of a sphere of radius 2024 and a sphere of radius 2023 displaced by √2, which is a circle, but the parts of the circle corresponding to smaller volumes also correspond to having another diagonal longer than 2024, and my minimum is the four diagonals that correspond to having two diagonals of length 2024.
But, as you point out, there are also diagonals to the intersection of a sphere of radius 2024 and a sphere of radius 1 displaced by √(2023²+1), giving arbitrarily small volumes.
With no puzzles this week, I decided to try last week's puzzle as applied to next year (2025).
1. Knowing that the rectangular prism with the largest volume for a given internal diagonal is a cube, I find that the sides of the cube should be 2025/√3 = 1169.134, the maximum volume (noninteger x,y,z) is (2025/√3)³ = 1598060439.6271. To find the closest volume to this maximum for integer x,y,z, I plotted a spreadsheet table where x steps down from 1169, y steps up from 1169, and the table is filled with the possible z values, z = √(2025² - x² - y²) and I look for integer values of z in the table. I find 2 candidates:
x * y * z = 1194 * 1158 * 1155 = 1596963060, and
x * y * z = 1183 * 1144 * 1180 = 1596955360
The first of these two is slightly bigger, so my answer to the first puzzle is:
x * y * z = 1194 * 1158 * 1155 = 1596963060 <<======================
2. 2025 has quite a long Collatz sequence, 157 numbers. A Collatz sequence that contains 2025 must therefore have more than 157 numbers. I started with numbers less than 2025 and worked down, looking for one with a Collatz sequence of more than 157 numbers. I looked only at those Collatz sequences longer than 157 and it appears to me that the smallest number with 2025 in its Collatz sequence is, ... 2025. A surprise considering the vast number of numbers in the Collatz sequences for numbers between 1 and 2025.
Taking another approach to the Collatz sequence puzzle, I considered what possible numbers could precede 2025 in a Collatz sequence. It appears that the preceding numbers must be 2025*2ⁿ for n=1,2,3...inf. Exponentially increasing numbers with never a number less than 2025. It appears that any year evenly divisible by 3 (like 2025) has the property that its Collatz sequence is not contained in the Collatz sequence of a lesser number.
Seems like the semicircle puzzle should have been on the 15th. Or on Star Wars Day.