Today I turned 27133 days old and one of this number's interesting properties is that it can be represented as a sum of two, three, four and five cubes and in more than one way in some cases. Specifically we can write: $$ \begin{align} 27133 &= 2744+24389\\&=14^3+29^3\\27133 &= 8+125+27000\\&=2^3+5^3+30^3\\27133&=1331+2744+3375+19683\\&=11^3+14^3+15^3+27^3\\27133&=8+8+1000+1728+24389\\&=2^3+2^3+10^3+12^3+29^3\\27133 &= 8+27+512+2197+24389\\&=2^3+3^3+8^3+13^3+29^3\\27133 &= 8+125+3375+8000+15625\\&=2^3+5^3+15^3+20^3+25^3\\27133 &= 125+4096+4913+5832+12167\\&=5^3+16^3+17^3+18^3+23^3\\27133&=216+3375+3375+8000+12167\\&=6^3+15^3+15^3+20^3+23^3 \end{align}$$This is all well and good but visually it's hardly exciting. What I'd like to be able to do is to create an octagonal balance with the eight cubes sitting at the ends of the eight arms. This would have some visual impact I'm sure.
Figure 1 |
Figure 2 |
Figures 1 and 2 are a start but how are the twin arms of the traditional scales in Figure 1 transformed into eight arms or four twin arms? How can all the cubes from \(2^3\) to \(30^3\) be represented in proportion, as is partially done in Figure2, and placed on the scales? It could be done with a lot of time and effort using traditional methods but how well will AI perform.
Here are the directions I gave to DALL-E and Figure 3 shows the four results:
Draw a set of scales with eight arms instead of the traditional two and have cubes of various sizes placed at the ends of the arms so that they roughly balance.
Figure 3 |
As can be seen the results are not impressive. Here is a second prompt that I gave it and Figure 4 shows the results:
Octopus balancing cubes of various sizes at the end of each of its tentacles.
Figure 4 |
Again the results are unimpressive but I tried again with the following prompt and figure 5 shows the results:
A weighing machine has eight arms formed from four intersecting metal rods. There are pans at the end of each arm on which to place items to be weighed. In this case, the pans contain a variety of multicolored cubes of various sizes that roughly balance one another.
Figure 5 |
Despite the very specific instructions the program wasn't up to the task. Figure 6 shows what NightCafe came up with using the same prompt as before:
Figure 6 |
I guess this is slightly better but it's still far short of what I want. Figure 7 is what starryai came up with:
Figure 7 |
Nope. DeepAI came up with the image shown in Figure 8.
Figure 8 |
Can you find an image of a scale for weighing that has eight trays instead of the traditional two, arranged evenly in a circle, so as to resemble the arms of an octopus.There is no image of a scale for weighing that has eight trays instead of the traditional two, arranged evenly in a circle, so as to resemble the arms of an octopus.The reason for this is that octopus scales are not designed for weighing. They are actually used for detecting motion and orientation. The eight arms of an octopus are covered in tiny suckers, and each sucker has a small sensor that can detect changes in pressure. This allows the octopus to sense the movement of its environment and to maintain its balance.A scale that has eight trays would be very difficult to use for weighing. The trays would have to be very small, and it would be difficult to keep them evenly balanced. Additionally, the scale would be very sensitive to vibrations, which would make it difficult to get accurate readings.So, while it is possible to imagine a scale that has eight trays, it is not practical or useful. Octopus scales are simply not designed for weighing.
However, Bard was more forthcoming with this straightforward request:
Can you find an image of an octahedron with the formula for its volume displayed.
Sure, I found you an image of an octahedron with the formula for its volume displayed.
The formula for the volume of an octahedron is:
Volume = (√2/3) * a^3
where a is the side length of the octahedron.
The octahedron is a three-dimensional shape with eight faces, each of which is an equilateral triangle. The volume of an octahedron is equal to one-third the volume of a regular tetrahedron with the same side length.
The site proved quite useful and so Bard can be used for finding mathematical content on the Internet, unlike ChatGPT 3.5 that does not have such access.
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