2025 AMC 12B Problem 23

Question

Let

S

be the set of all integers

z \gt 1

such that for all pairs of nonnegative integers

(x, y)

with

x \lt y \lt z,

the remainder when

2025x

is divided by

z

is less than the remainder when

2025y

is divided by

z.

What is the sum of the elements of

S?

3041

3542

3750

4044

4319

The condition says the map

k \mapsto 2025k \bmod z

is strictly increasing on

0, 1, \ldots, z-1

An increasing rearrangement of

0, 1, \ldots, z-1

starting at

0

must be the identity, forcing

2025 \equiv 1 \pmod z

The condition requires

k \mapsto 2025k \bmod z

to be strictly increasing on

\{0, 1, \ldots, z-1\}.

A strictly increasing list of

z

distinct values in

[0, z-1]

must be

0, 1, \ldots, z-1,

so

2025 \equiv 1 \pmod z,

i.e.

2024

is divisible by

z.

Since

2024 = 2^3 \cdot 11 \cdot 23,

the sum of all its divisors is

(1+2+4+8)(1+11)(1+23)

= 15 \cdot 12 \cdot 24

= 4320.

Excluding

z = 1

leaves

4319.

Thus, the correct answer is E.

Problem 23 in other years