# Mitosis of 80 bioelectronic cells

You are working in a bio-laboratory, where they produce some special bioelectronic cells which have mitosis ability.

FYI: Mitosis is a process of cell duplication during which one cell gives rise to two genetically identical daughter cells.

This very special cell divides itself into two cells by electronically controlled mitosis, which means it is controlled by a lab employee. In other words, lab employee commands the cell to divide into two. But after mitosis, cell size becomes half as well.

Your boss gave you $80$ of these electronic cells and want you to have the least amount of cells with the same size by mitosis.

By only applying mitosis to the cells, what is the least number of the same sized cells you can have?

• So the goal is to minimize the biggest group of cells of the same size? Or to minimize the total number of cells sharing a size with any other cell? – Deusovi May 24 '18 at 19:59
• @Deusovi yes, minimize the group of cells of the same size. – Oray May 24 '18 at 20:00
• Half the size in diameter, area or volume? – Phylyp May 24 '18 at 20:00
• @Phylyp - I don't think it matters. All that matters is that they become a different size, and we're trying to minimize the number of cells in the biggest group. – Deusovi May 24 '18 at 20:01
• @Phylyp as it does not matter to be honest, you can take it as area or volume or diameter as half, you just know which cell size group is different than others – Oray May 24 '18 at 20:01

41. Define the size of a cell that has undergone $k$ splits as $2^{-k}$. It is easy to see that the mitosis operation leaves total size invariant. Therefore, if we could get 40, our maximum size is $40(1+\frac12+\frac14+\frac18+\cdots)$ which by summing up the infinite series gives $80$, our original size. But presumably we are not allowed to use an infinite number of operations.
Now we demonstrate that 41 is possible. Keep 41 of the same size, split 39, now we have (41,78). Keep 41 of the smaller size, split 37, now we have (41,41,74). Once again - (41,41,41,66), and again (41,41,41,41,50), and again (41,41,41,41,41,18) and since $18\leq41$, we're done now!