Math
The modular clock
A dial of m marks and one hand. Adding a step draws a star that closes after m/gcd(s, m) moves; multiplying draws a cycle whose length divides phi(m). Some moduli have no primitive root at all, and the card says so - that is a theorem, not a failed search.
What happens
The hand moves one step at a time and the marks it touches light up; the ones that stay dark answer the question before the panel names the gcd. Adding 8 on a dial of 12 draws a triangle and lights three marks; adding 5 draws a star and lights all twelve, because 5 and 12 share no factor. The closed figure is the result, so it stays on screen instead of being wiped. Switch to multiplying and the rule changes: the hand goes k → k·g, the cycle length is the ORDER of g, and Euler's theorem says that order must DIVIDE phi(m) - on a dial of 17 the only lengths that can occur are 1, 2, 4, 8 and 16. A step whose order is exactly phi(m) is called a primitive root, and on a prime dial it walks every non-zero mark: that is the task of this card. How many such steps are there? Exactly phi(phi(m)) - two for 7, four for 11, ten for 23 - and the panel counts them by brute force as well, so the formula and the search agree in front of you. And then the part worth staying for: primitive roots do not exist for every modulus. They exist only for 1, 2, 4, p^k and 2p^k, so on a dial of 8, 12, 15, 16, 20, 21 or 24 there is none at all. The card says that out loud instead of quietly finding nothing, because 'there is no answer' is a different statement from 'I did not find one'.
Equations used in the module
Equations adding: the hand at position k moves to k + s (mod m); the cycle closes after m / gcd(s, m) moves and visits EVERY mark exactly when gcd(s, m) = 1 m = 12, s = 8 → 3 marks (a triangle); m = 12, s = 5 → all 12; m = 360, s = 84 → 30 multiplying: k → k·g (mod m); the cycle length is the ORDER of g, and by Euler's theorem it divides phi(m) m = 17: phi = 16 and the orders present are 1, 2, 4, 8, 16 - nothing else can occur a primitive root is a g whose order IS phi(m); for a prime m that means a cycle through all m − 1 non-zero marks how many: exactly phi(phi(m)) when they exist - m = 7 → 2 (smallest 3), m = 11 → 4 (smallest 2), m = 23 → 10 (smallest 5) WHEN THEY EXIST: only for m = 1, 2, 4, p^k and 2p^k. For 8, 12, 15, 16, 20, 21, 24 there is NO primitive root at all - a theorem, not a failed search phi is computed here, not tabulated: phi(7) = 6, phi(9) = 6, phi(12) = 4, phi(16) = 8, phi(36) = 12, phi(100) = 40, phi(1000) = 400
