In the game "Mathematical Clock," initially, a pointer points to one of the seven numbers contained in the clock's houses as illustrated in the figure below. In each round, the player must check which number the arrow points to and then move it, in a clockwise direction, by the number of spaces indicated by the number the arrow points to at the beginning of the round.

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-10.jpg?height=711&width=720&top_left_y=650&top_left_x=568)

For example, if the pointer initially points to the house with the number 2, the player must, in the first round, move it two units clockwise, so it will point to the house with the number 4.

a) Starting from the house with the number 1, how many rounds are necessary until the pointer returns to this house for the first time?

b) When starting the game with the pointer initially positioned at the house with the number 6, what will be its position after one round?

c) What is the only number from which, when starting the game, the arrow will point back to it in one round?

d) The player decides to switch the clock shown in the figure above for a clock containing 128 houses. Starting from the house with the number 127, how many rounds will be necessary for the pointer to reach the house with the number 128 for the first time?
Solution

a) Starting from house number 1, after the first round, the pointer should point to house number $1+1=2$. Performing another round, the pointer should move two houses clockwise, then pointing to house number $2+2=4$. On the third round, the pointer will be moved, starting from house number 4, four houses clockwise. This is equivalent to moving it by three houses and then one additional house. After moving three of these houses, the pointer will be positioned at house number $4+3=7$, and should now move one additional house clockwise. When it moves this additional house, the pointer will be positioned at house number 1, thus returning to its initial position. Therefore, the total number of steps required until the first return to the initial house is equal to three.
b) Starting from house number 6, the player should move the pointer six houses clockwise. This is equivalent to moving it, first by one house, and then by five additional houses. When moving it by one house, the pointer's position will be $6+1=7$. Now, when moving it by five more houses, its final position will be 5.

c) When starting the game from house number 7, the player should move the pointer seven houses clockwise, thus ending exactly at house number 7. From item a), we already know that starting from numbers 1, 2, and 4, the pointer does not return to the initial number in the first round. From item b), we also know that, when starting from number 6, the pointer will not point back to number 6 after one round. Starting from number 3, the pointer will point to number 6 after one round. Finally, starting from number 5, the pointer will point to house number 3 after one round. Thus, the house of number 7 is the only one that satisfies the property mentioned in the statement of this item.

d) Let $n=128$ and note that $127=n-1$. When starting the game from house number $n-1=127$, the pointer should move 127 houses clockwise. This is equivalent to moving one house to reach house number 128 and then moving 126 additional houses to reach, at the end, house number $n-2=126$. Thus, after one round, the pointer will have moved from house $n-1$ to house $n-2$.

In the second round, the pointer starts from house number $n-2=126$. Therefore, it should move two houses to reach house number 128 and then move 124 additional houses to reach, at the end, house number 124. Thus, at the end of the second round, the pointer will have moved from house number $n-2$ to house number $n-4$.

Repeating the same argument, we see that, when leaving house number $n-4$, the pointer should move to house number $n-8$ and from there to house number $n-16$, and so on until reaching house number $n-64$, which will occur after six rounds.

Note that $n-64=128-64=64$, so after six rounds, the pointer is at house number 64. From house number 64, the pointer moves 64 houses clockwise to reach house number $64+64=128$. Thus, the pointer reaches house number 128 for the first time after exactly seven rounds.

Observation: Note that $128=2^{7}$. The interested reader can generalize the solution to this problem to show that, if the clock has $n=2^{k}$ houses, then starting from house number $n-1$, house number $n$ will be reached for the first time after $k$ rounds.