C52 (41-4, Hungary) A magician has 100 cards, each labeled with a number from 1 to 100. He places these 100 cards into three boxes, one red, one white, and one blue. Each box must contain at least one card. A spectator then selects two of the three boxes, picks one card from each of the selected boxes, and announces the sum of the numbers on the two cards. Knowing this sum, the magician can identify which box was not selected. How many ways are there to place the cards so that the magician can always succeed? (Two methods are considered different if at least one card is placed in a different colored box.)


Consider the integers from 1 to 100. For simplicity, define the color of the box in which the integer $i$ is placed as the color of that integer. Use $r$ to represent red, $w$ to represent white, and $b$ to represent blue.
(1) There exists some $i$ such that the colors of $i, i+1, i+2$ are all different, for example, $r, w, b$. Since $i+(i+3)=(i+1)+(i+2)$, the color of $i+3$ cannot be the color of $i+1$, which is $w$, nor can it be the color of $i+2$, which is $b$. Therefore, it must be $r$. It is clear that if three consecutive numbers have different colors, the color of the next number can be determined. Furthermore, this color pattern will repeat: after $r w b$, it must be $r$, then $w, b, \cdots$ and so on. Similarly, this process holds in the reverse direction: before $r w b$, it must be $b, \cdots$ and so on.

Therefore, it is only necessary to determine the colors of $1, 2, 3$. There are 6 different ways to do this, and all 6 methods can make the magic trick successful, because their sums $r+w, w+b, b+r$ give different remainders modulo 3.
(2) There do not exist three consecutive numbers whose colors are all different. Assume 1 is red. Let $i$ be the smallest number that is not red, and assume $i$ is white. Let $k$ be the smallest blue number. By assumption, $i+11$ must be red, and by $t+99=(t-1)+100$, it follows that $t-1$ is blue, which contradicts the assumption that only 100 is blue.
Thus, the colors of these numbers must be $r w w \cdots w w b$. This method indeed works: if the sum of the numbers on the two selected cards is $\leqslant 100$, the box from which no card was selected must be blue;

if the sum of the numbers is $101$, the box from which no card was selected must be white;

if the sum of the numbers is > 101, the box from which no card was selected must be red.
Finally, there are 6 ways to arrange the colors in the above manner.
Therefore, there are 12 ways to place the cards so that the magic trick always succeeds.