AMC 10 · 2018 · #16

Grade 6 number-theory
Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue
modular-arithmeticexponentspattern-recognition easier-related-problempattern-recognition ↑ Prerequisites: modular-arithmetic
📏 Medium solution 💡 3 insights
Problem
Two thousand eighteen positive whole numbers a₁ ＜ a₂ ＜ … ＜ a₂018 add up to 2018²⁰¹⁸. Without knowing which numbers they are, find the remainder you get when a₁³ + a₂³ + … + a₂018³ is divided by 6.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

Tool #9 (Solve an Easier Related Problem): the question asks only for a remainder mod 6, so I can work the whole problem in remainders and never need the actual numbers. Tool #9 (Solve an Easier Related Problem): cubing 2018 unknown numbers looks impossible, but there is a remainder shortcut — every cube leaves the same remainder mod 6 as its base, so the hard sum of cubes collapses to the easy sum that the problem already hands me. Tool #5 (Look for a Pattern): the last step reduces to 2²⁰¹⁸ mod 6, and the powers of 2 fall into a short repeating cycle that pins the remainder instantly.

1STEP 1

A cube keeps its base's remainder mod 6

Compare a cube to its base: n³ − n = (n−1) n (n+1), a product of three consecutive whole numbers.

n³ - n = (n-1) n (n+1)
2STEP 2

Three in a row is divisible by 6

Three in a row hide a multiple of 2 and of 3, so the product is a multiple of 6 — hence n³ ≡ n (mod 6).

6 ∣ (n-1) n (n+1) → n³ ≡ n (mod 6)
3STEP 3

Swap the sum of cubes for the sum

Sum n³ ≡ n over all terms: the sum of cubes ≡ the given total 2018²⁰¹⁸ (mod 6) — the increasing, distinct details never matter.

Σ_i=1²⁰¹⁸ a_i³ ≡ Σ_i=1²⁰¹⁸ a_i = 2018²⁰¹⁸ (mod 6)
4STEP 4

Reduce the base 2018 mod 6

Reduce the base: 2018 = 6·336 + 2, so 2018 ≡ 2 (mod 6) and 2018²⁰¹⁸ ≡ 2²⁰¹⁸ (mod 6).

2018 = 6· 336 + 2 → 2018²⁰¹⁸ ≡ 2²⁰¹⁸ (mod 6)
5STEP 5

Powers of 2 cycle between 2 and 4

Powers of 2 mod 6 cycle 2, 4, 2, 4, …: an even exponent gives 4. As 2018 is even, 2²⁰¹⁸ ≡ 4 (mod 6), so the remainder is 4 — choice (E).

2²≡ 4, 2³≡ 2, 2⁴≡ 4,… → 2²⁰¹⁸ ≡ 4 (mod 6) → (E)
Answer
4
Sanity check the cube rule on small numbers: 1³=1≡1, 2³=8≡2, 3³=27≡3, 4³=64≡4, 5³=125≡5 — every cube matches its base mod 6, confirming n³≡ n. So the answer must equal 2018²⁰¹⁸mod 6. Cross-check that value: 2018²⁰¹⁸ is even, so its remainder mod 6 is even (0,2, or 4); and 2018≡ 2mod 3 gives 2018²⁰¹⁸≡ 2²⁰¹⁸≡ 1mod 3 (since 2²≡1 and the exponent is even), so the remainder is ≡1mod 3. The only value among 0,2,4 that is ≡1mod 3 is 4. Consistent with (E).
💡Key takeaway

Cubing never changes a number's remainder mod 6, so the messy sum of cubes has the same remainder as the given total 2018²⁰¹⁸, which works out to 4.

  • A cube keeps its base's remainder mod 6
  • Three in a row is divisible by 6
  • Swap the sum of cubes for the sum
  • Reduce the base 2018 mod 6
  • Powers of 2 cycle between 2 and 4

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