Problem 2 · Track a composition around a cycle · 3 points
Made by AI and not reviewed by a real person.
Let g move each number in {0, 1, …, 2025} one step forward around a circle, wrapping from 2025 to 0. How many permutations f have no fixed points and satisfy f(g(x)) = g⁻¹(f(x)) for every x?
- 2026
- 0
- 2
- 1013
- 1012
Show a hint
- Follow what the equation forces as you move from one number to the next around the circle.
- If a = f(0), express f(x) in terms of a, then ask when f(x) could equal x.
Show the answer and solution
Answer: D — 1013
Write a = f(0). Since applying g to an input moves it forward while applying g⁻¹ to an output moves it backward, the equation forces f(x) ≡ a − x (mod 2026). Every value of a gives a permutation. A fixed point would satisfy 2x ≡ a (mod 2026), which is possible exactly when a is even. There are 1013 odd values of a, so the answer is 1013. The tempting answer 2026 counts the permutations that do have fixed points as well.