🇺🇸 AMC 8 ⇄ switch contest
1997 AJHSME

Problem 19

Problem 19 · 1997 AJHSME Hard
Algebra & Patterns telescoping

If the product

32 · 43 · 54 · 65 · … · ab= 9,

what is the sum of a and b?

Show answer
Answer: D — 35.
Show hints
Hint 1 of 2
Don't multiply anything out — line up the fractions and watch the 3, then 4, then 5… appear on top of one fraction AND on the bottom of the next. They annihilate in a chain.
Still stuck? Show hint 2 →
Hint 2 of 2
This is a telescoping product: in 3/2 · 4/3 · 5/4 · … every numerator cancels the following denominator, collapsing the whole chain to (last top) / (first bottom).
Show solution
Approach: telescoping cancellation
  1. Each numerator equals the next fraction's denominator (the 3 on top of 3/2 cancels the 3 under 4/3, and so on), so the entire product collapses to a/2 — only the very last numerator and the very first denominator survive.
  2. Set a/2 = 9 → a = 18. Since each fraction is (top)/(top − 1), b = a − 1 = 17.
  3. Sum = 18 + 17 = 35.
  4. Why this transfers: whenever consecutive terms share a factor that cancels its neighbor, the long product (or sum) telescopes to just the endpoints — recognizing the pattern saves all the multiplying.
Mark: · log in to save