23505 and the Visualisation of Triprimes

Yesterday I was 23505 days old. This number is triprime with factors of 2, 3 and 1567. I'm not sure how widely the term "triprime" is used in the mathematical community but it follows on logically enough from the term "biprime". One way of visualising a triprime number is to associate it with a rectangular prism whose length, breadth and height correspond to the three factors. Suppose we want to create a rectangular prism with a volume of 23505 cubic metres using only square plates with sides of one metre. There is only one way to do this and that is with a prism whose sides measure 2, 3 and 1567 metres (or 1.567 kilometres).

Viewed in one way, this prism has a very narrow cross-section of 2 metres by 3 metres and it is very inefficient in terms of the surface areas required to enclose the volume. The most efficient shape would be a cube with a side of \(23505^{1/3}\) metres. This cube would have a surface area of \(6 \times 23505^{2/3} \) square metres as opposed to our prism with a surface area of$$(2 \times 3 + 3 \times 1567 + 2 \times 1567) \times 2 \text{ m}^2 $$As was done with biprimes earlier, we could then determine the percentage "surface area efficiency" of our prism according to the formula:$$ \text{efficiency}=\frac{\text{surface area of cube}}{ \text{surface area of rectangular prism}} \times 100$$In the case of 23505, the efficiency turns out to be a little over 31.39% by my calculations.

So biprimes can be visualised as unique rectangles (that in some cases can be squares) and triprimes can be visualised as unique rectangular prisms (that in some cases can be cubes). Tetraprimes and beyond of course can have no visualisation in 3-dimensional space.

23501 and the Visualisation of Biprimes

Today I'm 23501 days old. This number is biprime, meaning that it has two prime factors (71 and 331). One way to get a handle on understanding the significance of biprimes is to imagine that you have an area of 23501 to enclose with a rectangular fence made up of modular 1 metre sections. This means of course that both sides of the rectangle must be whole numbers. There is only one way to do this and that is by creating a rectangle with a length of 331 metres and a width of 71 metres.

With other numbers that are neither prime nor biprime (such as 23502), there is more than one way. For example, because 23502 = 2 x 3 x 3917, there are three possible rectangles:

• 2 metres by 11751 metres representing a perimeter of 23.506 kilometres
• 3 metres by 7834 metres representing a perimeter of 15.674 kilometres
• 6 metres by 3917 metres representing a perimeter of 7.846 kilometres

These are admittedly very narrow rectangles but rectangles none the less. Of course, they are very inefficient in terms of the perimeter length required to cover the required area. Without the restraint of whole numbers, the most efficient four sided figure would be a square with a side of a little more than 153.3 metres and perimeter of about 613.2 metres or 0.6132 kilometres.

You could measure, in percentage terms, the efficiency of the perimeter required to enclose the area by using the formula:$$ \text{efficiency}=\frac{ \text{perimeter of square}}{ \text{perimeter of rectangle}} \times 100$$In this case, the efficiencies become:

• 2.6%
• 3.9%
• 7.8%

Getting back to 23501 however, the efficiency of the rectangle turns out to be 76.3%. Of course, with more factors the number of possible rectangles increase and rectangles of greater "perimeter efficiency" can be defined. Of course, biprimes with equal factors will convert to squares and thus have 100% efficiency.