Problem 21 · 1990 AJHSME
Stretch
Algebra & Patterns
work-backward
A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three numbers are 16, 64, 1024.
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Answer: B — 1/4.
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Hint 1 of 2
You're given the END of the list and asked for the START — that's a signal to run the rule *backward*. The forward rule is 'multiply the two before,' so the reverse is a division. Which two numbers do you divide?
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Hint 2 of 2
If a term equals (term before it) × (term two before it), then 'two before' = (this term) ÷ (the one just before). Keep peeling backward one step at a time until you reach the first number.
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Approach: work backward — undo each multiply with a divide
- Label the list t₁…t₈; the rule is tₙ = tₙ₋₁ × tₙ₋₂. We know t₆=16, t₇=64, t₈=1024 (and indeed 16×64 = 1024 ✓, a good consistency check).
- To go back, rearrange the rule: tₙ₋₂ = tₙ ÷ tₙ₋₁. Step down: t₅ = t₇÷t₆ = 64÷16 = 4; t₄ = t₆÷t₅ = 16÷4 = 4; t₃ = t₅÷t₄ = 4÷4 = 1; t₂ = t₄÷t₃ = 4÷1 = 4; t₁ = t₃÷t₂ = 1÷4.
- So the first number is 1/4. (Verify forward: 1/4, 4 → 1, 4, 4, 16, 64, 1024 ✓.)
- *Why this transfers:* when a problem hands you the end of a chain, reverse the operation (multiply→divide, add→subtract) and walk back — always sanity-check by running it forward.
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