AMC 10 · 2024 · #18

Grade 8 arithmetic
modular-arithmeticexponentseulers-theoremprime-factorization complementary-countingidentify-subproblemscasework ↑ Prerequisites: modular-arithmeticexponentsprime-factorization
📏 Medium solution 💡 3 insights
Problem
Let n range over every integer. How many distinct remainders can the expression n¹⁰⁰ mod 125 take?

Pick an answer.

(A)
1
(B)
2
(C)
5
(D)
25
(E)
125

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Every integer either shares the prime 5 with 125 or it does not — Tool #16 (Change Focus) reframes the problem by that dichotomy and turns one hard question into two easy ones. Tool #7 (Identify Subproblems) then handles each case: (a) n coprime to 125 (no factor of 5) — here Euler's theorem nails n¹⁰⁰ ≡ 1 (mod 125) in one line because φ(125) = 100; (b) n is a multiple of 5 — then n¹⁰⁰ = 5¹⁰⁰ k¹⁰⁰, divisible by 5¹⁰⁰ ≫ 5³ = 125, so the remainder is 0. Tool #9 (Solve an Easier Related Problem) checks the coprime conclusion on a small case (n = 2) before trusting the general theorem, so the answer is grounded, not hand-waved.

1STEP 1

125 = , so only the prime 5 matters: every integer n is either coprime to 125 or a multiple of 5.

integers = {n : 5 ∤ n} n : 5 ∣ n
2STEP 2

Case A — 5 ∤ n: since φ(125) = 100 and the exponent is 100, Euler's theorem gives n¹⁰⁰ ≡ 1 (mod 125).

φ(125) = 5³ - 5² = 100, n¹⁰⁰ ≡ 1 (mod 125)
3STEP 3

Check n = 2: 2¹⁰ ≡ 24 and 24⁵ ≡ -1, so 2¹⁰⁰ ≡ (-1)² = 1 (mod 125) — Case A confirmed.

2¹⁰ ≡ 24, 24⁵ ≡ -1, 2¹⁰⁰ ≡ (-1)² = 1 (mod 125)
4STEP 4

Case B — 5 ∣ n: write n = 5m, then n¹⁰⁰ = 5¹⁰⁰ m¹⁰⁰ is a multiple of 5³ = 125, so the remainder is 0.

n¹⁰⁰ = 5¹⁰⁰ m¹⁰⁰ = 125 · 5⁹⁷ m¹⁰⁰ ≡ 0 (mod 125)
5STEP 5

Every integer n lands in exactly one case, so the possible remainders are just {0, 1} — 2 distinct values → (B).

{n¹⁰⁰ mod 125 : n ∈ Z} = {0, 1}, |{0,1}| = 2 → (B)
Answer
2
Both remainders are attained: n = 5 gives 5¹⁰⁰, divisible by 125, remainder 0; n = 1 gives 1¹⁰⁰ = 1, remainder 1. Both values are in {0, 1, …, 124} and lie in the legal remainder range. The five answer choices include 1, 2, 5, 25, 125 — exactly the small divisor-like numbers one would expect from a guess; {0, 1} has size 2, matching (B). Quick sanity for Case A on another base: n = 3: 3⁵ = 243 ≡ 243 - 125 = 118 ≡ -7 (mod 125), then 3¹⁰ ≡ 49, 3²⁰ ≡ 49² = 2401 = 19 · 125 + 26 ≡ 26, 3⁵⁰ ≡ 26² · 49 ≡ 676 · 49 ≡ 51 · 49 = 2499 ≡ -1, so 3¹⁰⁰ ≡ 1 (mod 125). Same answer, confirming Case A.
💡Key takeaway

This AMC 10 problem only needs the Grade 8 integer-exponent rule and a clean "is n a multiple of 5 or not?" split — and the answer set turns out to be just {0, 1}!