Counting the Reachable Marks
Lesson 03 answered which marks the pointer can reach. This lesson asks how many. On the 5-, 7- and 11-mark dials the answers were 3, 4 and 6. You will find the pattern, discover which rotations land on the same mark, and prove the count for every odd prime, without listing any readings.
Where this lesson is headed
In words: the rotations pair up, two to a mark, except for one rotation that is alone. So p rotations cover 1 + (p − 1)/2 marks. Same setup as Lesson 03: the clock starts with circumference p, grows 1 cm per rotation, and is read on a p-mark dial, so the reading is T(k) mod p. At the end you will see the same count holds for the growing clock with any starting circumference and growth.
p: number of dial marks (an odd prime)
k: rotations completed, 0 to p − 1
T(k): 0 + 1 + ⋯ + (k − 1) = k(k − 1)/2
Reading: T(k) mod p
Reachable marks: each mark counted once, however many rotations land there
01. How many marks get reached?
Start with the 5-mark dial. Turn the clock through all p rotations and count the marks that light up. Then try bigger dials and look for a formula. Pay attention to which rotations share a mark: that is the key to the whole lesson. (Shortcut: Space turns one rotation, ← undoes one.)
The p-mark dial
Rotation k → mark
Blue chips land on a new mark; orange chips land on a mark already used. Count each mark once.
Settings
Dial size p (odd primes)
Which marks have been reached?
| k | T(k) | Reading | Also reached by |
|---|
Predict before you finish
Guided experiments
Do these in order. "Set this up" chooses the dial and turns the clock for you.
02. Which rotations land on the same mark?
If we can say exactly when two rotations give the same reading, the count follows. Each step has the general idea on the left and your numbers on the right.
Two rotations a and b land on the same mark
We want to find every b that does this for a given a.
Clear the fraction and move everything to one side
Multiply by 2. On an odd dial this can be undone (multiply by "half", (p + 1)/2, from Lesson 03), so nothing is lost:
Factor
Expand and regroup, using a² − b² = (a − b)(a + b):
So the question becomes: when is (a − b)(a + b − 1) a multiple of p?
Use that p is prime
A prime that divides a product must divide one of the factors (Euclid's lemma). So either p divides a − b, or p divides a + b − 1:
The first option is the same rotation. The second is its partner: b = 1 − a, which on the dial is p + 1 − a. These are the mirror pairs from Lesson 03, and now we know there are no other collisions.
This is the step that needs a prime. On a 9-mark dial, 3 × 3 = 9 is a multiple of 9 without either factor being one, and extra collisions appear (Stage 03).
Pair up all the rotations
Send each k to its partner 1 − k. Doing it twice gives back k, so the rotations split into pairs. The only rotation left alone is one that is its own partner:
On an odd dial "half" exists and is unique (Lesson 03), so there is exactly one lone rotation.
Count the groups
The p rotations form 1 lone rotation and (p − 1)/2 pairs. Each group lands on one mark, and by step 4 different groups land on different marks. So
03. Prove the count, and see how far it reaches
There are two proofs: one by pairing rotations (Stage 02), and one by counting squares (Lesson 03). Then we find where the count fails, and see that it covers far more than T(k).
Theorem · odd prime p
Proof workshop
Complete proof · pairing
For an odd prime p, T(a) ≡ T(b) is equivalent (multiply or divide by 2) to (a − b)(a + b − 1) ≡ 0. Since p is prime, this holds exactly when b ≡ a or b ≡ 1 − a. So the rotations landing on any one mark are exactly a group {a, 1 − a}.
Applying k ↦ 1 − k twice gives k again, so each group has two members unless k ≡ 1 − k, that is 2k ≡ 1, which has exactly one solution because p is odd. Hence the p rotations split into one lone rotation and (p − 1)/2 pairs, and different groups land on different marks. The number of marks is 1 + (p − 1)/2 = (p + 1)/2. ∎
The lone rotation k = (p + 1)/2 is the only one landing on its mark. In the square test of Lesson 03, that mark is the one whose 8a + 1 reads 0.
Second proof · count the squares
From Lesson 03:
On an odd dial, k ↦ 2k − 1 hits every mark exactly once, and so does r ↦ 8r + 1 (both can be undone). So the number of reachable marks equals the number of squares on the dial.
Squares pair up the same way: y and −y (that is, p − y) have the same square, and nothing else does, because y² ≡ z² means (y − z)(y + z) ≡ 0, and p is prime. The only y that is its own partner is 0.
Where does the count fail?
Try dials that break one of the assumptions and compare (m + 1)/2 with the actual count.
Odd non-prime dials break step 4 (extra collisions). Even dials break step 5 (no "half").
The bigger picture
The whole growing clock, with any N and g
In Lesson 02 the 5-mark dial reached 3 marks whether N was 1 or 3. Here is why. Read the full total S(k) = kN + gT(k) on a p-mark dial. Written out,
which is a quadratic in k. The same folding works for any quadratic Ak² + Bk + C on a prime dial, as long as A does not read 0: completing the square turns it into A(k + shift)² + constant, and shifting, scaling by a non-zero number and adding a constant never change how many marks are hit. For the clock, A = g/2 reads 0 exactly when g is a multiple of p. So:
Try any quadratic Ak² + Bk + C
| k | f(k) | Reading | New or repeat |
|---|
Computer check
Checks every odd prime up to 97: the count, that exactly one rotation is alone and the rest come in pairs, and that the lone rotation is (p + 1)/2. Also checks many growing clocks and general quadratics on small prime dials.
Exit challenge · explain it in your own words
Why does the 13-mark dial reach exactly 7 marks, and why does the same reasoning work for every odd prime? Your explanation should state (a) when two rotations share a mark, (b) where "prime" is used, and (c) why exactly one rotation is alone.
What a complete answer includes
Derive (a − b)(a + b − 1) ≡ 0; use that a prime dividing a product divides a factor, so the partner of k is 1 − k; show k ↦ 1 − k undoes itself and has the single fixed point 2k ≡ 1, i.e. k = 7 on 13 marks; count one lone rotation plus (p − 1)/2 pairs. For p = 13 that is 1 + 6 = 7.