2. In each part of this problem, cups are arranged in a circle and numbered $1,2,3$, and so on. A ball is placed in cup 1. Then, moving clockwise around the circle, a ball is placed in every $n$th cup. The process ends when cup 1 contains two balls. For example, starting with 12 cups and placing a ball in every 3rd cup, balls are placed, in order, in cups $1,4,7,10,1$.
(a) There are 12 cups in the circle and a ball is placed in every 5 th cup, beginning and ending with cup 1. List, in order, the cups in which the balls are placed.
(b) There are 9 cups in the circle and a ball is placed in every 6th cup, beginning and ending with cup 1 . List the numbers of the cups that do not receive a ball.
(c) There are 120 cups in the circle and a ball is placed in every 3rd cup, beginning and ending with cup 1. How many cups do not contain at least one ball when the process is complete? Explain how you obtained your answer.
(d) There are 1000 cups in the circle and a ball is placed in every 7 th cup, beginning and ending with cup 1 . Determine the number of the cup into which the 338 th ball is placed.
2. (a) The 1st ball goes in cup 1 .
Each successive ball goes 5 cups further along the sequence. We obtain the cup number for the next ball by adding 5 to the current cup number. If the total is no larger than 12 , this is the cup number of the next ball; if the cup number is larger than 12 , we subtract 12 to obtain the actual cup number.
Thus, the 2 nd ball goes in cup $1+5=6$, the 3 rd ball goes in cup $6+5=11$, and the 4 th ball goes in cup " $11+5=16$ " which is $16-12=4$ cups beyond cup 12 , so is really cup 4 . We continue until we place a second ball in cup 1 .
In order, the balls are placed in cups $1,6,11,4,9,2,7,12,5,10,3,8,1$.
(b) The first ball goes in cup 1.
The next ball goes in cup $1+6=7$.
The next ball goes in cup $7+6=13$, which is the same as cup $13-9=4$.
The next ball goes in cup $4+6=10$, which is the same as cup $10-9=1$.
This means that the process stops.
Therefore, balls are placed in 3 cups in total (cups 1, 7,4) and so cups 2, 3, 5, 6, 8, and 9 do not receive a ball.
(c) The first ball goes in cup 1 .
Cup 1 will be reached again after moving 120 cups around the circle.
Since a ball is placed in every 3 rd cup, then 120 cups corresponds to $\frac{120}{3}=40$ additional balls, and the 41st ball will be placed in cup 1 and the process stops.
Therefore, there will be 40 cups in which a ball is placed. (Cup 1 will receive two balls.) Since 40 cups receive a ball, then $120-40=80$ cups do not contain at least one ball.
(d) The first ball goes in cup 1 .
Since each ball is placed 7 cups further along the circle, then the 338 th ball will be placed $337 \times 7=2359$ cups further along the circle from cup 1 .
Cup 1 itself is 1000 cups further along the circle, so we can think of moving 1000 cups to cup 1 , then another 1000 cups to cup 1 again, and then 359 further cups to cup 360 .
Therefore, the 338th ball is placed in cup 360 .