Problem 17 · 2019 AMC 8
Hard
Fractions, Decimals & Percents
fraction-to-decimal
What is the value of the product
(1·32·2)(2·43·3)(3·54·4) … (97·9998·98)(98·10099·99) ?
Show answer
Answer: B — 50/99.
Show hints
Hint 1 of 2
98 factors is a signal: this is meant to collapse, not be multiplied out. Each fraction is k(k+2) over (k+1)2 — numbers one apart, top and bottom — which begs to be split so neighbors cancel.
Still stuck? Show hint 2 →
Hint 2 of 2
Each factor k(k+2)(k+1)(k+1) breaks into kk+1 × k+2k+1; collect all the first pieces in one chain and all the second pieces in another.
Show solution
Approach: split each factor into two telescoping chains
- Every factor is k(k+2)(k+1)(k+1) = kk+1 × k+2k+1, for k = 1 to 98.
- Chain 1 (the kk+1 pieces): 12 × 23 × … × 9899. Each top cancels the next bottom, leaving 199.
- Chain 2 (the k+2k+1 pieces): 32 × 43 × … × 10099, which cancels down to 1002 = 50.
- Multiply the two leftovers: 199 × 50 = 5099.
- Why this transfers: a long product or sum that looks hopeless is usually telescoping — each piece cancels part of its neighbor. Factor every term into simple pieces, line them up, and almost everything collapses, leaving just the two ends.
Mark:
· log in to save