Problem 22 · 2023 AMC 8
Hard
Number Theory
factorizationsubstitution
In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is 4000. What is the first term?
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Answer: D — 5.
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Hint 1 of 2
Don't guess numbers — track structure. Call the first two terms a and b; every later term is a product, so it's just a and b raised to some powers. Watch how those powers grow.
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Hint 2 of 2
The exponents of a (and of b) each follow a Fibonacci-style add: the 6th term is a3b5. Now break 4000 into primes and match.
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Approach: track exponents of a and b through the sequence
- Let the first two terms be a, b. Each new term multiplies the previous two, which adds their exponents — so the exponents climb like Fibonacci: a, b, ab, ab2, a2b3, a3b5.
- So the 6th term is a3b5 = 4000. Factor: 4000 = 4 × 1000 = 22 × (2×5)3 = 25 × 53. Matching the cube to a3 and the fifth power to b5 gives a = 5, b = 2.
- First term: 5. Sanity check: the sequence reads 5, 2, 10, 20, 200, 4000 — the 6th term hits 4000. Worth keeping: in any ‘multiply the previous two’ sequence, the exponents of the two starting values follow the Fibonacci numbers.
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