The +1 Bonus Theorem
In Lesson 05 the 1st block on a power-of-two dial skipped the halfway mark, and the 2nd block filled it in. But in Lessons 03 to 05 the growth was always 1 cm. Now let the growth g be anything. A clock with starting circumference N is read on an N-mark dial, and we count how many marks N rotations reach. Sometimes there is exactly one extra mark, a "bonus". You will find out exactly when.
Where this lesson is headed
In words: the gcd shrinks the dial from N marks to m marks, and N rotations are d blocks of m rotations. The 1st block reaches B(m) marks. Later blocks add exactly one more mark when there is a later block (d > 1) and m is a power of two, and nothing otherwise.
N: starting circumference, and the number of dial marks
g: growth per rotation (g ≥ 0)
d = gcd(N, g), m = N/d: the smaller dial
B(m): marks T(k) mod m reaches for k = 0 to m − 1 (the base count)
C(N, g): marks the clock reaches on the N-mark dial in N rotations
01. Look for one extra mark
Start where Lesson 05 ended: an 8-mark dial with growth g = 1. Eight rotations reach 7 marks. Now change the growth to 2 and watch what happens. The small dial on the right shows the same readings shrunk to m = N/gcd(N, g) marks, as in Lesson 02, step 6. (Shortcut: Space turns one rotation, ← undoes one.)
The N-mark dial
Settings
N · starting circumference and dial marks
Small dial: T(k) mod m
Big dial: reading on N marks
Guided experiments
Do these in order. "Set this up" chooses N and g and turns the clock for you.
Rotation by rotation
| k | T(k) | Big dial: gT(k) mod N | Small dial: T(k) mod m | Block | New? |
|---|
Your conjecture
How many extra marks can the later blocks add?
Compare the cases side by side
These are the experiments above in one table. Which property does the bonus row have that the others do not?
| N, g | d | m | B(m) | C(N, g) | Extra |
|---|
02. Shrink the dial, then count the blocks
Every piece of this argument is something you have already proved. Each step has the general idea on the left and your numbers on the right.
On an N-mark dial only the growth part shows
The total is S(k) = kN + gT(k). On a dial with N marks, kN is whole trips (Lesson 02, step 5):
Shrink the dial by the gcd
Let d = gcd(N, g) and write N = dm, g = du, so u and m share no factor. Lesson 02, step 6 proved
So two rotations share a mark on the big dial exactly when they share one on the small m-mark dial. Counting marks on the big dial is the same as counting T(k) mod m.
Dividing the modulus by d is what makes this legal; cancelling u then works because u and m share no factor.
N rotations are d blocks of m
k = 0 to N − 1 splits into d blocks of m rotations each. The 1st block (k = 0 to m − 1) reaches a set of marks on the small dial; call it A. It has B(m) marks.
If d = 1 there is only one block, so C(N, g) = B(m). No bonus is possible.
Each block is the one before, turned by the jump
From Lesson 02, over m rotations the small-dial reading moves by the jump:
which reads 0 if m is odd and m/2 if m is even. Two jumps always cancel. So every block reaches either A or A turned by m/2, and
Four cases
m = 1: the small dial has one mark, so C = 1 = B(1).
m odd: the jump is 0, later blocks repeat A exactly. C = B(m).
m = 2, 4, 8, …: Lesson 05 says A is every mark except m/2. Turned by m/2 it is every mark except 0. Together: all m marks. C = m = B(m) + 1. The bonus.
m even, not a power of two (6, 10, 12, …): turning A by m/2 gives A back again, so nothing is added. This is the "no-bonus lemma"; Lesson 07 proves it. Use the tool on the right to see it.
Put it together
Both conditions matter: without a 2nd block (d = 1) the bonus has no chance to appear, and m = 1 is excluded because a 1-mark dial has nothing to add.
Check the edge case
For N = 12, g = 12 we get d = 12 and m = 1. How many marks are reached?
03. Prove the count in every case
The proof combines Lesson 02 (shrinking the dial and the jump), Lesson 05 (the skipped halfway mark), and one fact proved in Lesson 07, the no-bonus lemma for even dials that are not powers of two.
The theorem
N ≥ 1 and g ≥ 0, with gcd(N, 0) = N.
Complete proof
- Let d = gcd(N, g), m = N/d, u = g/d. On the N-mark dial the reading is gT(k) mod N (Lesson 02, step 5). Since u and m share no factor, two rotations share a mark on the N-mark dial exactly when T(a) ≡ T(b) (mod m) (Lesson 02, step 6). So C(N, g) is the number of values of T(k) mod m for k = 0 to N − 1.
- These N rotations are d blocks of m. The 1st block reaches a set A with B(m) marks. If d = 1, C = B(m).
- If m = 1, everything reads 0 and C = 1 = B(1). From now on d > 1 and m ≥ 2.
- T(k + m) − T(k) = mk + m(m − 1)/2. If m is odd this is a multiple of m, so every block reaches A and C = B(m).
- If m is even, let h = m/2. Then T(k + m) ≡ T(k) + h and 2h ≡ 0 (mod m), so the blocks reach exactly A ∪ (A + h).
- If m = 2r, Lesson 05 gives A = every mark except h, so A + h = every mark except 0, and the union is all m marks: C = m = B(m) + 1.
- If m is even but not a power of two, the no-bonus lemma gives A + h = A, so C = B(m). The cases cover every possibility. □
Note: step 7 really needs the lemma. For a general set, containing h does not make it unchanged when turned by h; for T it does, because of mirror pairs, and Lesson 07 proves this.
Looking ahead · why dials like 6, 10, 12 add nothing
The no-bonus lemma is proved in Lesson 07, in two moves. First, mirror pairs: T(1 − k) = T(k), so every 2nd-block rotation except k = m and m + 1 has a partner in the 1st block, and those two both read m/2. So the 2nd block can add a mark only if m/2 is missing from the 1st block. Second, when m = 2rs with s odd and s > 1, the Chinese Remainder Theorem finds a rotation inside the 1st block that reads m/2. For powers of two (s = 1) no such rotation exists, which is exactly the bonus.
Computer check
Checks every pair 1 ≤ N ≤ 150 and 0 ≤ g ≤ N by listing the readings directly and comparing with the formula, and separately checks the no-bonus lemma for every even m up to 150. Many matches are strong evidence, but only the proof covers every case.
Exit problem · show your work
Find C(48, 6) and C(30, 5). Work out d, m and B(m) and decide whether the bonus applies; then check by turning the clock in Stage 01 if you like.