Competition · AMC preparation · step 4 of 4

AMC 8 · 2019 · #17

Grade 5 arithmeticpattern

Archetype Telescoping Collapse └ Sub-archetype Product Ratio Cancellation

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

\frac{(k · (k+2))}{(k+1)}

· (k+1), starting at k = 1 (so the first factor is

\frac{(1 · 3)}{(2 · 2)}

) and ending at k = 98 (so the last factor is

\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)

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) · (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

Try just two factors

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

(1 · 3)/(2 · 2) · (2 · 4)/(3 · 3) = (1 · 3)(2 · 4)/(2 · 2)(3 · 3) = 24/36 = 2/3 · 1/2 = (1 · 4)/(2 · 3) · 1/1

2STEP 2

Redo the small product cleanly

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

2 factors: (1 · 4)/(2 · 3) = 4/6; 3 factors: (1 · 5)/(2 · 4) = 5/8

3STEP 3

Read off the pattern

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

n=2: (1 · 4)/(2 · 3), n=3: (1 · 5)/(2 · 4), n=4: (1 · 6)/(2 · 5), …, n: (1 · (n+2))/(2 · (n+1))

4STEP 4

Split and telescope

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

Chain 1 = 1/99, Chain 2 = 100/2 = 50

5STEP 5

Combine the two chains

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

P = 1/99 · 50 = 50/99 → (B)

Answer

50/99

Each individual factor

\frac{(k(k+2))}{((k+1)²)}

= 1 -

\frac{1}{((k+1)²)}

is just slightly less than 1 (e.g.,

\frac{(1 · 3)}{4}

= 0.75,

\frac{(2 · 4)}{9}

≈ 0.889,

\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

\frac{50}{99}

≈ 0.505 — roughly

\frac{1}{2}

— fits perfectly. Answer (A)

\frac{1}{2}

is dangerously close but the formula gives exactly

\frac{50}{99}

, not

\frac{1}{2}

(since

\frac{50}{99}

≠

\frac{49.5}{99}

). Answer (C)

\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!

Try just two factors
Redo the small product cleanly
Read off the pattern
Split and telescope
Combine the two chains

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Try just two factors

Redo the small product cleanly

Read off the pattern

Split and telescope

Combine the two chains

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