AMC 10 · 2010 · #16

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
Three numbers are drawn at random and an expression built from them must be a multiple of three. Find the probability.

Pick an answer.

(A)
$\dfrac{1}{3}$
(B)
$\dfrac{29}{81}$
(C)
$\dfrac{31}{81}$
(D)
$\dfrac{11}{27}$
(E)
$\dfrac{13}{27}$
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

The range splits evenly into three remainders.

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

Factor the expression

Factoring pulls one number out front.

abc + ab + a = a (bc + b + 1)
3STEP 3

Handle the easy case: a is a multiple of 3

If that one is a multiple, it works always.

a ≡ 0: always divisible, P = 1/3
4STEP 4

List all remainder cases for b and c

Otherwise only two remainder pairs survive.

{(b,c): bc+b+1 ≡ 0} = {(1,1), (2,0)}, P = 2/9
5STEP 5

Combine the two cases

Combining gives 13/27, choice (E).

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

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