AMC 10 · 2023 · #24

Grade 8 number-theory
prime-factorizationgcdlcmp-adic-valuationcasework identify-subproblemscaseworkextreme-principle ↑ Prerequisites: gcdlcmprime-factorization
📏 Long solution 💡 5 insights
Problem
Four positive integers have a given product, and the least common multiple of each of the six pairs among them is given too. Find the greatest common divisor of all four.

Pick an answer.

(A)
30
(B)
45
(C)
3
(D)
15
(E)
6
How to solve
Strategy Identify Subproblems

Seven multiplicative conditions tangle four unknown numbers together, and there is no useful place to start while they stay whole numbers. Tool #7 (Identify Subproblems) is the whole move: switch from the numbers to their prime exponents, and the seven conditions fall apart into three puzzles that never mention each other, one for the twos, one for the threes, one for the fives. Tool #4 (Introduce a Variable) makes that switch possible by naming the twelve exponents. Tool #15 (Organize Information in More Ways) supplies the dictionary that does the translating: multiplying adds exponents, an lcm takes the larger exponent, a gcd takes the smaller. Inside each small puzzle, tool #14 (Extreme Principle) starts at the tightest condition, because the lowest ceiling is the one that pins other numbers down. Tool #3 (Eliminate Possibilities) then removes a capped number from contention whenever a high maximum has to be reached by somebody. Tool #2 (Make a Systematic List) closes each prime by sweeping the handful of exponent quadruples that survive, which also proves the answer does not depend on which one is chosen.

1STEP 1

Turn them into exponents

Turn each into three exponents.

a = 2^a₂ 3^a₃ 5^a₅, b = 2^b₂ 3^b₃ 5^b₅, c = 2^c₂ 3^c₃ 5^c₅, d = 2^d₂ 3^d₃ 5^d₅
2STEP 2

Translate the operations

Products add; the lcm takes the larger.

nu_p(xy) = nu_p(x) + nu_p(y), nu_p(lcm(x,y)) = max(nu_p(x), nu_p(y)), nu_p(gcd(x,y)) = min(nu_p(x), nu_p(y))
3STEP 3

Three primes, three puzzles

Each prime is a separate puzzle.

prime 2: sum 6, prime 3: sum 9, prime 5: sum 7, gcd(a,b,c,d) = 2^m₂ · 3^m₃ · 5^m₅
4STEP 4

Solve the first prime

Start from the lowest ceiling.

max(b₂,c₂)=1 → b₂ ≤ 1, c₂ ≤ 1 → d₂ = 2, a₂ = 3; b₂ + c₂ = 1 → (a₂,b₂,c₂,d₂) = (3,1,0,2) or (3,0,1,2), m₂ = 0
5STEP 5

Solve the second prime

A low cap forces the values.

max(a₃,b₃)=2 → a₃ ≤ 2, b₃ ≤ 2 → c₃ = d₃ = 3; a₃ + b₃ = 3 → (a₃,b₃,c₃,d₃) = (1,2,3,3) or (2,1,3,3), m₃ = 1
6STEP 6

Solve the third prime

The third works the same way.

b₅, c₅, d₅ ≤ 2 → a₅ = 3; at least two of b₅, c₅, d₅ equal 2; b₅ + c₅ + d₅ = 4 → {b₅, c₅, d₅} = {2, 2, 0}, m₅ = 0
7STEP 7

Rebuild the answer prime by prime

Rebuilding gives 3.

m₂ = 0, m₃ = 1, m₅ = 0 ⟹ gcd(a,b,c,d) = 2⁰ · 3¹ · 5⁰ = 3
Answer
3
Build an actual quadruple and test all seven conditions. Take a₂,a₃,a₅ = 3,1,3 and so on from one arrangement per prime: a = 2³ 3¹ 5³ = 3000, b = 2¹ 3² 5² = 450, c = 2⁰ 3³ 5² = 675, d = 2² 3³ 5⁰ = 108. Their product is 98415000000 = 2⁶ · 3⁹ · 5⁷. The six lcms come out lcm(a,b) = 9000 = 2³ 3² 5³, lcm(a,c) = lcm(a,d) = 27000 = 2³ 3³ 5³, lcm(b,c) = 1350 = 2¹ 3³ 5², and lcm(b,d) = lcm(c,d) = 2700 = 2² 3³ 5², matching every line of the problem. And gcd(3000, 450, 675, 108) = 3, which is choice (C). A full sweep of each prime confirms the counts used above: prime 2 has exactly 2 valid quadruples, prime 3 exactly 2, prime 5 exactly 3, and the minima are 0, 1, 0 in every case, so no choice of arrangement changes the answer. The four wrong choices are exactly the ones that smuggle a factor back in: 30 and 6 need an even gcd, but one of b, c has no factor of 2 at all; 30, 15, and 45 need a factor of 5, but one of b, c, d has none.
💡Key takeaway

When a problem is built out of multiplying, lcm and gcd, stop looking at the numbers and count prime factors instead: each prime becomes its own small puzzle, and lcm just means take the bigger count.

  • Turn each number into three exponents
  • Multiplying adds, lcm takes the larger
  • Three primes, three separate puzzles
  • Prime 2: the tightest ceiling first
  • Prime 3: one low cap forces two threes
  • Prime 5: two of the three must be 2
  • Rebuild the gcd prime by prime