AMC 8 · 2022 · #8
Easy mode Grade 5Problem
Imagine a long line of fractions, all multiplied together. The first fraction is 31. The next is 42. The next is 53. 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,… all the way up to 20. So the last fraction is 2220.
What do you get when you multiply all of these fractions together?
31⋅42⋅53⋯2018⋅2119⋅2220?
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $\tfrac{1}{22}$
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
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 $\frac{1\cdot 2}{4\cdot 5}$.
💡 Working a tiny version first turns the scary $20$-factor product into a $3$-factor product we can actually compute.
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.
💡 The same shape pops out — only the first two numerators and the last two denominators survive. That is the pattern to ride.
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.
💡 Cancelling the same factor from top and bottom does not change a fraction — it just makes it shorter.
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 $\frac{2}{462}$.
💡 Multiplying two small fractions is a Grade 5 "fraction times fraction" calculation.
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.
💡 Reducing $\tfrac{2}{462}$ by dividing both by $2$ uses the Grade 4 idea that an equivalent fraction has the same value.
5.NF.B.4 Try a much shorter version of the same product to feel the pattern. Take just th 4.OA.C.5 Try one more easier case to confirm the pattern. Take the first four fractions: 4.NF.A.1 State the general pattern. Writing the full product as a single fraction gives $ 5.NF.B.4 Multiply out the tiny survivors. Top: $1\times 2 = 2$. Bottom: $21\times 22 = 46 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.5Generate 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.1Explain 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.4Apply 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 $\tfrac{1\cdot 2}{21\cdot 22}$!
⭐ 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 $\tfrac{1\cdot 2}{21\cdot 22}$!
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