AMC 8 · 2022 · #8

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

fraction-multiplicationpattern-recognitionfraction-arithmetic pattern-recognitionidentify-subproblems ↑ Prerequisites: fraction-multiplicationfraction-arithmetic

📏 Short solution 💡 2 insights

📘 View easy version →

Problem

Find the value of the long product

\frac{1}{3}

\frac{2}{4}

\frac{3}{5}

…

\frac{18}{20}

\frac{19}{21}

\frac{20}{22}

, where the k-th factor is

\frac{k}{k+2}

and k runs from 1 to 20. Pick the matching choice from (A)

\frac{1}{462}

, (B)

\frac{1}{231}

, (C)

\frac{1}{132}

, (D)

\frac{2}{213}

, (E)

\frac{1}{22}

Pick an answer.

(A)

$frac{1}{462}$

(B)

$frac{1}{231}$

(C)

$frac{1}{132}$

(D)

$frac{2}{213}$

(E)

$frac{1}{22}$

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Multiplying out 20 fractions by brute force would give a 20-digit numerator and denominator — wildly impractical for AMC 8 timing, and a clear signal to look for structure. Tool #5 (Look for a Pattern) spots that each numerator k (for k ≥ 3) reappears as the denominator of the fraction two steps earlier (since denominator k comes from $\frac{k-2}{k}$ ), so almost everything cancels. To make that pattern concrete first, Tool #9 (Solve an Easier Related Problem) tries the same product with just 3 or 4 fractions, watches what survives, and generalizes. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net at the end.

1STEP 1

Warm up on the first three fractions $\frac{1}{3}$ · $\frac{2}{4}$ · $\frac{3}{5}$ : the 3 on top cancels the 3 on the bottom, leaving $\frac{1·2}{4·5}$ = $\frac{1}{10}$ .

\frac{1· 2· 3}{3· 4· 5}

\frac{1· 2}{4· 5}

\frac{2}{20}

\frac{1}{10}

2STEP 2

Add a fourth fraction $\frac{1}{3}$ · $\frac{2}{4}$ · $\frac{3}{5}$ · $\frac{4}{6}$ : the 3 and 4 both cancel, so only 1·2 up top and 5·6 below survive, giving $\frac{1}{15}$ .

\frac{1· 2· 3· 4}{3· 4· 5· 6}

\frac{1· 2}{5· 6}

\frac{2}{30}

\frac{1}{15}

3STEP 3

In the full product every integer 3 to 20 cancels, leaving only 1·2 on top and 21·22 on the bottom: $\frac{1·2}{21·22}$ .

\frac{1· 2· 3· 4… 20}{3· 4… 20· 21· 22}

\frac{1· 2}{21· 22}

4STEP 4

Multiply the survivors: 1×2 = 2 on top, 21×22 = 462 below, so the product is $\frac{2}{462}$ .

\frac{1· 2}{21· 22}

\frac{2}{462}

5STEP 5

Divide top and bottom by 2 to reduce $\frac{2}{462}$ to $\frac{1}{231}$ , which is choice (B) — none of the other choices equal $\frac{2}{462}$ .

\frac{2}{462}

\frac{2÷ 2}{462÷ 2}

\frac{1}{231}

→ (B)

Answer

frac{1}{231}

Every factor

\frac{k}{k+2}

is less than 1, and we multiply 20 of them, so the answer must be a very small positive fraction — exactly what

\frac{1}{231}

is. Also, an alternate check: the unreduced form

\frac{2}{462}

has 462 = 2· 3· 7· 11 = 21· 22, which matches the "two leftover denominators are 21 and 22" structure we found. (A)

\frac{1}{462}

is exactly half as big — that would be the mistake of forgetting that the top survivor is 1· 2 = 2, not 1. (E)

\frac{1}{22}

is 21 times too big — that would be the mistake of forgetting that 21 also stays in the denominator.

💡Key takeaway

This AMC 8 problem only needs Grade 5 fraction multiplication you already know — spot the cancellation pattern with a tiny version first, and the scary 20-fraction product shrinks to $\frac{1· 2}{21· 22}$ !

More like this