AMC 8 · 2019 · #17

Grade 5 arithmeticpattern
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
fraction-multiplicationpattern-recognitionsequences-arithmetic pattern-recognitionidentify-subproblems ↑ Prerequisites: fraction-multiplicationpattern-recognition
📏 Medium solution 💡 3 insights
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Problem
We multiply 98 fractions of the form (k⋅(k+2))(k+1)\frac{(k · (k+2))}{(k+1)} · (k+1), starting at k = 1 (so the first factor is (1⋅3)(2⋅2)\frac{(1 · 3)}{(2 · 2)}) and ending at k = 98 (so the last factor is (98⋅100)(99⋅99)\frac{(98 · 100)}{(99 · 99)}). What single number does this big product equal?

Pick an answer.

(A)
$\frac{1}{2}$
(B)
$\frac{50}{99}$
(C)
$\frac{9800}{9801}$
(D)
$\frac{100}{99}$
(E)
50

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

The product has 98 factors — way too many to multiply by hand. Tool #9 (Easier Related Problem) says: try the same product with only 2, then 3, then 4 factors first, watch what happens, and conjecture a formula. Tool #5 (Look for a Pattern) reads the small-case results into a clean rule. Tool #7 (Identify Subproblems) gives the cleanest shortcut: split each factor as k(k+1)\frac{k}{(k+1)} · (k+2)(k+1)\frac{(k+2)}{(k+1)}, which turns the giant product into two telescoping chains that each collapse in one line. Both routes give the same answer; the small-case path is friendlier for a young solver, and the telescoping split confirms it.

1STEP 1

Start with the easiest version — just the first two factors (k = 1, 2), multiplying tops together and bottoms together.

(1⋅3)(2⋅2)\frac{(1 · 3)}{(2 · 2)} · (2⋅4)(3⋅3)\frac{(2 · 4)}{(3 · 3)} = (1⋅3)(2⋅4)(2⋅2)(3⋅3)\frac{(1 · 3)(2 · 4)}{(2 · 2)(3 · 3)} = 2436\frac{24}{36} = 23\frac{2}{3} · 12\frac{1}{2} = (1⋅4)(2⋅3)\frac{(1 · 4)}{(2 · 3)} · 11\frac{1}{1}
2STEP 2

The two-factor product cleans up to 23\frac{2}{3}; now redo it with the first three factors (k = 1, 2, 3) and watch the same numbers cancel.

2 factors: (1⋅4)(2⋅3)\frac{(1 · 4)}{(2 · 3)} = 46\frac{4}{6}; 3 factors: (1⋅5)(2⋅4)\frac{(1 · 5)}{(2 · 4)} = 58\frac{5}{8}
3STEP 3

List the small cases in a row: only the end numbers survive, giving (1⋅(n+2))(2⋅(n+1))\frac{(1 · (n+2))}{(2 · (n+1))} for n factors.

n=2: (1⋅4)(2⋅3)\frac{(1 · 4)}{(2 · 3)}, n=3: (1⋅5)(2⋅4)\frac{(1 · 5)}{(2 · 4)}, n=4: (1⋅6)(2⋅5)\frac{(1 · 6)}{(2 · 5)}, …, n: (1⋅(n+2))(2⋅(n+1))\frac{(1 · (n+2))}{(2 · (n+1))}
4STEP 4

Split each factor into k(k+1)\frac{k}{(k+1)} · (k+2)(k+1)\frac{(k+2)}{(k+1)}; the two telescoping chains collapse to 199\frac{1}{99} and 50.

Chain 1 = 199\frac{1}{99}, Chain 2 = 1002\frac{100}{2} = 50
5STEP 5

Multiply the two chains: 199\frac{1}{99} · 50 = 5099\frac{50}{99} — matching the small-case formula at n = 98, so the answer is (B).

P = 199\frac{1}{99} · 50 = 5099\frac{50}{99} → (B)
Answer
5099\frac{50}{99}
Each individual factor (k(k+2))((k+1)2)\frac{(k(k+2))}{((k+1)²)} = 1 - 1((k+1)2)\frac{1}{((k+1)²)} is just slightly less than 1 (e.g., (1⋅3)4\frac{(1 · 3)}{4} = 0.75, (2⋅4)9\frac{(2 · 4)}{9} ≈ 0.889, (3⋅5)16\frac{(3 · 5)}{16} ≈ 0.9375, climbing toward 1). Multiplying 98 numbers each less than 1 must give something less than 1, and the early factors drag it well below 1. The value 5099\frac{50}{99} ≈ 0.505 — roughly 12\frac{1}{2} — fits perfectly. Answer (A) 12\frac{1}{2} is dangerously close but the formula gives exactly 5099\frac{50}{99}, not 12\frac{1}{2} (since 5099\frac{50}{99} ≠ 49.599\frac{49.5}{99}). Answer (C) 98009801\frac{9800}{9801} is too close to 1, (D) and (E) are larger than 1, so they cannot be right.
💡Key takeaway

This AMC 8 problem only needs Grade 5 fraction multiplication you already know — try a few small cases, spot the pattern, and the giant product solves itself!

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