AMC 8 · 2022 · #8

Easy mode Grade 5

Problem

Imagine a long line of fractions, all multiplied together. The first fraction is $\frac{1}{3}$ . The next is $\frac{2}{4}$ . The next is $\frac{3}{5}$ . The pattern keeps going.

In each fraction, the bottom number is $2$ more than the top number. The top numbers run through $1, 2, 3, \dots$ all the way up to $20$ . So the last fraction is $\frac{20}{22}$ .

What do you get when you multiply all of these fractions together?

$\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?$

$\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}$

Pick an answer.

(A)

$frac{1}{462}$

(B)

$frac{1}{231}$

(C)

$frac{1}{132}$

(D)

$frac{2}{213}$

(E)

$frac{1}{22}$

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Toolkit + CCSS Solution

Understand

Restated: Find the value of the long product $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}$, where the $k$-th factor is $\frac{k}{k+2}$ and $k$ runs from $1$ to $20$. Pick the matching choice from (A) $\tfrac{1}{462}$, (B) $\tfrac{1}{231}$, (C) $\tfrac{1}{132}$, (D) $\tfrac{2}{213}$, (E) $\tfrac{1}{22}$.

Givens: There are $20$ fractions multiplied together; The $k$-th fraction is $\frac{k}{k+2}$ for $k = 1, 2, \ldots, 20$; Numerators are the integers $1, 2, 3, \ldots, 20$; Denominators are the integers $3, 4, 5, \ldots, 22$; Answer choices: (A) $\tfrac{1}{462}$, (B) $\tfrac{1}{231}$, (C) $\tfrac{1}{132}$, (D) $\tfrac{2}{213}$, (E) $\tfrac{1}{22}$

Unknowns: The simplified value of the $20$-fraction product

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

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 $k \geq 3$) 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.

Execute — Answer: B

#9 Solve an Easier Related Problem 5.NF.B.4 Step 1

Try a much shorter version of the same product to feel the pattern.
Take just the first three fractions: $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}$.
Multiply tops together and bottoms together.
The numerator $3$ matches a $3$ in the denominator, so it cancels, leaving $\frac{1\cdot 2}{4\cdot 5}$.

$$\dfrac{1\cdot 2\cdot 3}{3\cdot 4\cdot 5} = \dfrac{1\cdot 2}{4\cdot 5} = \dfrac{2}{20} = \dfrac{1}{10}$$

💡 Working a tiny version first turns the scary $20$-factor product into a $3$-factor product we can actually compute.

#5 Look for a Pattern 4.OA.C.5 Step 2

Try one more easier case to confirm the pattern.
Take the first four fractions: $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdot\frac{4}{6}$.
The $3$ on top cancels the $3$ on the bottom, and the $4$ on top cancels the $4$ on the bottom — only the two smallest numerators ($1, 2$) and the two largest denominators ($5, 6$) survive.

$$\dfrac{1\cdot 2\cdot 3\cdot 4}{3\cdot 4\cdot 5\cdot 6} = \dfrac{1\cdot 2}{5\cdot 6} = \dfrac{2}{30} = \dfrac{1}{15}$$

💡 The same shape pops out — only the first two numerators and the last two denominators survive. That is the pattern to ride.

#5 Look for a Pattern 4.NF.A.1 Step 3

State the general pattern.
Writing the full product as a single fraction gives $\frac{1\cdot 2\cdot 3\cdots 20}{3\cdot 4\cdot 5\cdots 22}$.
Every integer from $3$ to $20$ appears in both lists, so they cancel one-for-one.
What remains on top is the $1\cdot 2$ that was "too small" to also appear on the bottom, and what remains on the bottom is the $21\cdot 22$ that was "too big" to also appear on top.

$$\dfrac{1\cdot 2\cdot \cancel{3\cdot 4\cdots 20}}{\cancel{3\cdot 4\cdots 20}\cdot 21\cdot 22} = \dfrac{1\cdot 2}{21\cdot 22}$$

💡 Cancelling the same factor from top and bottom does not change a fraction — it just makes it shorter.

#5 Look for a Pattern 5.NF.B.4 Step 4

Multiply out the tiny survivors.
Top: $1\times 2 = 2$.
Bottom: $21\times 22 = 462$.
So the product equals $\frac{2}{462}$.

$$\dfrac{1\cdot 2}{21\cdot 22} = \dfrac{2}{462}$$

💡 Multiplying two small fractions is a Grade 5 "fraction times fraction" calculation.

#3 Eliminate Possibilities 4.NF.A.1 Step 5

Reduce to lowest terms by dividing top and bottom by their common factor $2$.
That gives $\frac{1}{231}$, which matches answer choice (B).
A quick scan of the other choices ($\tfrac{1}{462}$, $\tfrac{1}{132}$, $\tfrac{2}{213}$, $\tfrac{1}{22}$) shows none of them equal $\tfrac{2}{462}$, so (B) is the only match.

$$\dfrac{2}{462} = \dfrac{2\div 2}{462\div 2} = \dfrac{1}{231} \;\Rightarrow\; \textbf{(B)}$$

💡 Reducing $\tfrac{2}{462}$ by dividing both by $2$ uses the Grade 4 idea that an equivalent fraction has the same value.

[1] #9 5.NF.B.4 Try a much shorter version of the same product to feel the pattern. Take just th

[2] #5 4.OA.C.5 Try one more easier case to confirm the pattern. Take the first four fractions:

[3] #5 4.NF.A.1 State the general pattern. Writing the full product as a single fraction gives $

[4] #5 5.NF.B.4 Multiply out the tiny survivors. Top: $1\times 2 = 2$. Bottom: $21\times 22 = 46

[5] #3 4.NF.A.1 Reduce to lowest terms by dividing top and bottom by their common factor $2$. Th

Review

Reasonableness: 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 $\tfrac{1}{231}$ is. Also, an alternate check: the unreduced form $\tfrac{2}{462}$ has $462 = 2\cdot 3\cdot 7\cdot 11 = 21\cdot 22$, which matches the "two leftover denominators are $21$ and $22$" structure we found. (A) $\tfrac{1}{462}$ is exactly half as big — that would be the mistake of forgetting that the top survivor is $1\cdot 2 = 2$, not $1$. (E) $\tfrac{1}{22}$ is $21$ times too big — that would be the mistake of forgetting that $21$ also stays in the denominator.

Alternative: Tool #13 (Convert to Algebra) gives a clean closed form: the product is $\prod_{k=1}^{20} \frac{k}{k+2} = \frac{20!}{(22!/(2!))} = \frac{2!\cdot 20!}{22!} = \frac{2}{21\cdot 22} = \frac{1}{231}$. It is more compact but hides the cancellation intuition that #5 and #9 expose, so it is a verification path rather than the first-choice teaching path.

CCSS standards used (min grade 5)

4.OA.C.5 Generate a number or shape pattern following a given rule (Computing the $3$-factor and $4$-factor warm-up products and noticing the rule "only the first two numerators and the last two denominators survive".)
4.NF.A.1 Explain why a fraction is equivalent to another fraction (Cancelling the matched block $3\cdot 4\cdots 20$ from numerator and denominator and reducing $\tfrac{2}{462}$ to $\tfrac{1}{231}$ using the equivalent-fractions idea.)
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying the small-case products $\tfrac{1\cdot 2\cdot 3}{3\cdot 4\cdot 5}$ and computing the final survivor $\tfrac{1\cdot 2}{21\cdot 22}$ by multiplying numerators and denominators.)

⭐ 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 $\tfrac{1\cdot 2}{21\cdot 22}$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 5)

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