B4. A hundred elves (female scouts) are sitting in a large circle around the campfire. Each elf has one or more chestnuts, and no two elves have the same number of chestnuts. Each elf divides her number by the number of her right neighbor and writes the remainder on a green note. Each elf also divides her number by the number of her left neighbor and writes that remainder on a red note. For example, if Anja has 23 chestnuts and her right neighbor Bregje has 5, then Anja writes 3 on her green note and Bregje writes 5 on her red note.

If the number of different remainders on the 100 green notes is equal to 2, what is the smallest possible number of different remainders on the 100 red notes?
B4. 100 We number the gnomes in order around the circle. The gnome with the smallest number of chestnuts is given number 1. Her left neighbor is gnome 2, the left neighbor of gnome 2 is gnome 3, and so on up to gnome 100, who is the right neighbor of gnome 1. The number of chestnuts of gnome $k$ we call $a_{k}$.

We now look at the two neighbors of gnome 1. The right neighbor has $a_{100}$ chestnuts. Since $a_{1}a_{1}$, dividing $a_{2}$ by $a_{1}$ gives a remainder $r$ that is smaller than $a_{1}$. This remainder $r$ is on the green slip of gnome 2. The two numbers that appear on the hundred green slips are thus the two numbers $a_{1}$ and $r$.

We now look at gnomes 1 to 99. We prove that $a_{1}<a_{2}<\cdots<a_{100}$. Suppose this were not the case. For some $k$ it would then hold that $a_{k+1}<a_{k}$ (equality cannot hold because everyone has a different number of chestnuts). The remainder of $a_{k+1}$ when divided by $a_{k}$ is thus $a_{k+1}$ and that is the number on the green slip of gnome $k+1$. This number must therefore be equal to $a_{1}$ or to $r$. But that is impossible, because $a_{1}<a_{k+1}$ (gnome 1 has the smallest number of chestnuts) and $r$ is even smaller than $a_{1}$. We conclude that indeed $a_{1}<a_{2}<\cdots<a_{100}$.

If we now look at the numbers on the red slips of gnomes 1 to 99, we see that these are the numbers $a_{1}$ to $a_{99}$. The number on the red slip of gnome 100 is the remainder when $a_{100}$ is divided by $a_{1}$. This remainder is smaller than $a_{1}$, so the hundred numbers on the red slips are all different.