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2025 AMC 12B Problem 23

Problem 23 of 25HarderNumber Theory

Let SS be the set of all integers z>1z \gt 1 such that for all pairs of nonnegative integers (x,y)(x, y) with x<y<z,x \lt y \lt z, the remainder when 2025x2025x is divided by zz is less than the remainder when 2025y2025y is divided by z.z. What is the sum of the elements of S?S?

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Solution

The condition requires k↦2025k mod zk \mapsto 2025k \bmod z to be strictly increasing on {0,1,…,z−1}.\{0, 1, \ldots, z-1\}. A strictly increasing list of zz distinct values in [0,z−1][0, z-1] must be 0,1,…,z−1,0, 1, \ldots, z-1, so 2025≡1(modz),2025 \equiv 1 \pmod z, i.e. 20242024 is divisible by z.z. Since 2024=23⋅11⋅23,2024 = 2^3 \cdot 11 \cdot 23, the sum of all its divisors is (1+2+4+8)(1+11)(1+23)(1+2+4+8)(1+11)(1+23) =15⋅12⋅24= 15 \cdot 12 \cdot 24 =4320.= 4320. Excluding z=1z = 1 leaves 4319.4319. Thus, the correct answer is E.

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Concepts: modular arithmetic · divisibility · sum of factors

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.