AMC 10 · 2010 · #18

Grade 7 probabilitynumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

modular-arithmeticprobability-basicpolynomial-factoring casework ↑ Prerequisites: probability-basicmodular-arithmetic

📏 Medium solution 💡 3 insights

Problem

Pick three numbers a, b, c independently at random, each equally likely to be any of 1, 2, 3, …, 2010. Find the probability that abc + ab + a is a multiple of 3.

Pick an answer.

(A)

$\dfrac{1}{3}$

(B)

$\dfrac{29}{81}$

(C)

$\dfrac{31}{81}$

(D)

$\dfrac{11}{27}$

(E)

$\dfrac{13}{27}$

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Divisibility by 3 depends only on remainders mod 3, so Tool #9 (Solve an Easier Related Problem) shrinks the giant set {1,…,2010} down to just three equally likely remainders 0, 1, 2. Tool #7 (Identify Subproblems) factors the expression into a(bc + b + 1), splitting the question into 'is a a multiple of 3?' and 'is bc + b + 1 a multiple of 3?'. Tool #2 (Make a Systematic List) then counts the tiny finite table of remainder cases exactly, with no guessing.

1STEP 1

Only remainders mod 3 matter

Divisibility by 3 depends only on the remainder, and 2010 = 3 × 670, so each remainder 0, 1, 2 has probability 1/3.

P(a ≡ 0) = P(a ≡ 1) = P(a ≡ 2) = 1/3 (mod 3)

2STEP 2

Factor the expression

Every term shares a factor of a, so abc + ab + a = a(bc + b + 1), and a product is a multiple of 3 as soon as one factor is.

abc + ab + a = a (bc + b + 1)

3STEP 3

Handle the easy case: a is a multiple of 3

If a ≡ 0 (mod 3) the product works whatever b and c are — probability 1/3; otherwise we need bc + b + 1 ≡ 0.

a ≡ 0: always divisible, P = 1/3

4STEP 4

List all remainder cases for b and c

Across the nine remainder pairs, bc + b + 1 ≡ 0 only for (b, c) = (1, 1) and (2, 0), a chance of 2/9.

{(b,c): bc+b+1 ≡ 0} = {(1,1), (2,0)}, P = 2/9

5STEP 5

Combine the two cases

The two cases now exclude each other, so add: 1/3 + 2/3 × 2/9 = 9/27 + 4/27 = 13/27.

1/3 + 2/3·2/9 = 9/27 + 4/27 = 13/27

Answer

13/27

The result 13/27 ≈ 0.481 is a bit under one half, which is sensible: divisibility by 3 alone would give 1/3, and the extra factor structure only adds chances, so the answer should exceed 1/3 but stay well below 1. Cross-check by direct counting of remainder triples: over all 3³ = 27 equally likely remainder patterns for (a,b,c), the 9 with a ≡ 0 all work, and among the 18 with a ≢ 0 exactly the ones with (b,c) ∈ {(1,1),(2,0)} work, giving 2 × 2 = 4 more. Total 9 + 4 = 13 out of 27, confirming 13/27.

💡Key takeaway

For divisibility by 3, only remainders mod 3 matter — factor the expression, then check the tiny table of remainder cases instead of the huge original set.

Only remainders mod 3 matter
Factor the expression
Handle the easy case: a is a multiple of 3
List all remainder cases for b and c
Combine the two cases

Only remainders mod 3 matter

Factor the expression

Handle the easy case: a is a multiple of 3

List all remainder cases for b and c

Combine the two cases

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