AMC 10 · 2025 · #23

Grade 7 number-theory
modular-arithmeticdivisor-sumprime-factorization extreme-principleconvert-to-algebra ↑ Prerequisites: modular-arithmetic
📏 Long solution 💡 4 insights
Problem
Pick an integer z > 1. Multiply each of 0, 1, 2, …, z-1 by 2025 and record the remainder after dividing by z. Call z good when those remainders always come out in increasing order: for every pair x < y below z, the remainder from 2025x is smaller than the remainder from 2025y. Find every good z and add them all together.

Pick an answer.

(A)
3041
(B)
3542
(C)
3750
(D)
4044
(E)
4319
How to solve
Strategy Extreme Principle

The condition names a huge pile of pairwise comparisons, which is hopeless to check one pair at a time. So I first collect all the remainders into a single list and ask what an increasing list can even look like. That is where the whole problem cracks: the list has exactly as many entries as there are available remainders, so it is squeezed with zero slack and is forced to be 0, 1, 2, …, z-1 in order. From there the work turns algebraic — the forced equality becomes a divisibility statement about 2024 — and the leftover job is the routine subproblem of factoring 2024 and adding its divisors.

1STEP 1

Collect the remainders into one list

Collect the remainders into one list.

R(x) = 2025x mod z, R(0) < R(1) < … < R(z-1)
2STEP 2

The list has no room to skip

The slots fit exactly, so each remainder equals x.

R(0) < R(1) < … < R(z-1) inside {0, 1, …, z-1} → R(x) = x
3STEP 3

Turn the forced list into divisibility

That says z divides 2024.

2025x ≡ x (mod z) ⇔ z ∣ 2024x, x = 1 → z ∣ 2024
4STEP 4

Check the reverse direction

The converse also holds.

z ∣ 2024 → 2025 ≡ 1 (mod z) → R(x) = x for 0 ≤ x < z
5STEP 5

Break 2024 into primes

2024 is eight times eleven times 23.

2024 = 2³ · 11 · 23, 4 · 2 · 2 = 16 divisors
6STEP 6

Add the divisors, then drop 1

The divisor sum minus one is 4319.

(1 + 2 + 4 + 8)(1 + 11)(1 + 23) = 15 · 12 · 24 = 4320, 4320 - 1 = 4319
Answer
4319
Spot-check both sides of the rule. For z = 2, a divisor of 2024, the list is R(0) = 0 and R(1) = 1, which increases. For z = 3, not a divisor of 2024, R(0) = 0 and R(1) = 0 because 2025 is a multiple of 3, so the list fails at its very first step. Listing the survivors directly gives 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012, 2024, which is 15 numbers — exactly the 16 divisors minus the excluded 1 — and adding them by hand also gives 4319. The total also sits one below 4320, the full divisor sum of 2024, which is the exact shape the argument predicts.
💡Key takeaway

When an increasing list has exactly as many entries as there are values it can use, it has no room to skip anything and must be 0, 1, 2 and so on in order — and that squeeze turns a scary rule about remainders into the plain question of what divides 2024.

  • Collect the remainders into one list
  • The list has no room to skip
  • Turn the forced list into divisibility
  • Check the reverse direction
  • Break 2024 into primes
  • Add the divisors, then drop 1