AMC 8 · 2018 · #2
Grade 5 arithmeticpatternProblem
What is the value of the product
(1+11)⋅(1+21)⋅(1+31)⋅(1+41)⋅(1+51)⋅(1+61)?
Pick an answer.
AMC 8 2018 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: Compute the value of the product $\left(1+\tfrac{1}{1}\right)\left(1+\tfrac{1}{2}\right)\left(1+\tfrac{1}{3}\right)\left(1+\tfrac{1}{4}\right)\left(1+\tfrac{1}{5}\right)\left(1+\tfrac{1}{6}\right)$ and match the result to one of the five answer choices.
Givens: A product of six factors, each of the form $1 + \tfrac{1}{k}$ for $k = 1, 2, 3, 4, 5, 6$; Answer choices: (A) $\tfrac{7}{6}$, (B) $\tfrac{4}{3}$, (C) $\tfrac{7}{2}$, (D) $7$, (E) $8$
Unknowns: The exact value of the full six-factor product
Understand
Restated: Compute the value of the product $\left(1+\tfrac{1}{1}\right)\left(1+\tfrac{1}{2}\right)\left(1+\tfrac{1}{3}\right)\left(1+\tfrac{1}{4}\right)\left(1+\tfrac{1}{5}\right)\left(1+\tfrac{1}{6}\right)$ and match the result to one of the five answer choices.
Givens: A product of six factors, each of the form $1 + \tfrac{1}{k}$ for $k = 1, 2, 3, 4, 5, 6$; Answer choices: (A) $\tfrac{7}{6}$, (B) $\tfrac{4}{3}$, (C) $\tfrac{7}{2}$, (D) $7$, (E) $8$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem
Each factor $1 + \tfrac{1}{k}$ rewrites cleanly as $\tfrac{k+1}{k}$, so the product becomes $\tfrac{2}{1}\cdot\tfrac{3}{2}\cdot\tfrac{4}{3}\cdots\tfrac{7}{6}$ — a chain where each numerator matches the next denominator. Tool #5 (Look for a Pattern) is exactly the move that spots this telescoping cancellation, turning a six-fraction multiplication into a one-step answer. Tool #9 (Easier Problem) backs it up: trying the same product with only $2$ or $3$ factors first reveals the rule "the answer is just the last numerator" before we trust it on all six.
Execute — Answer: D
4.NF.B.3 Step 1 - Rewrite each factor as a single fraction.
- For any whole number $k$, $1 + \tfrac{1}{k} = \tfrac{k}{k} + \tfrac{1}{k} = \tfrac{k+1}{k}$, so the six factors become $\tfrac{2}{1}, \tfrac{3}{2}, \tfrac{4}{3}, \tfrac{5}{4}, \tfrac{6}{5}, \tfrac{7}{6}$.
💡 Adding a whole number and a unit fraction with the same denominator is just combining unit fractions — a Grade 4 fraction skill.
4.OA.C.5 Step 2 - Try the pattern on smaller cases first (Tool #9).
- With $2$ factors: $\tfrac{2}{1}\cdot\tfrac{3}{2} = 3$.
- With $3$ factors: $\tfrac{2}{1}\cdot\tfrac{3}{2}\cdot\tfrac{4}{3} = 4$.
- With $4$ factors: $\tfrac{2}{1}\cdot\tfrac{3}{2}\cdot\tfrac{4}{3}\cdot\tfrac{5}{4} = 5$.
- The pattern is clear: every numerator cancels with the next denominator, leaving just the last numerator.
💡 Generating a pattern from a few simple cases and stating its rule is the Grade 4 "shape/number pattern" standard.
5.NF.B.4 Step 3 - Apply the telescoping cancellation to the full six-factor product.
- Each numerator $2, 3, 4, 5, 6$ cancels with the matching denominator in the next factor, leaving only the very first denominator ($1$) and the very last numerator ($7$).
💡 Multiplying fractions and cancelling common factors top-and-bottom is exactly the Grade 5 fraction-by-fraction multiplication standard.
4.NF.B.3 Step 4 - Read off the value.
- $\tfrac{7}{1} = 7$, which matches choice (D).
💡 A fraction with denominator $1$ is just the numerator — a Grade 4 fraction-meaning idea.
4.NF.B.3 Rewrite each factor as a single fraction. For any whole number $k$, $1 + \tfrac{ 4.OA.C.5 Try the pattern on smaller cases first (Tool #9). With $2$ factors: $\tfrac{2}{1 5.NF.B.4 Apply the telescoping cancellation to the full six-factor product. Each numerato 4.NF.B.3 Read off the value. $\tfrac{7}{1} = 7$, which matches choice (D). Review
Reasonableness: Each factor is greater than $1$, so the product must be greater than $1$ — that already eliminates nothing, but the smallest factor is $\tfrac{7}{6} \approx 1.17$ and the largest is $2$, so a rough estimate $2 \times 1.5 \times 1.33 \times 1.25 \times 1.2 \times 1.17 \approx 7$ matches the exact answer. The pattern check is even stronger: with $n$ factors of the form $1+\tfrac{1}{k}$ for $k=1\ldots n$, the product equals $n+1$. Here $n = 6$, so the answer is $7$ — choice (D).
Alternative: Tool #3 (Eliminate Possibilities) on the choices: choices (A) $\tfrac{7}{6}$ and (B) $\tfrac{4}{3}$ are smaller than the very first factor ($2$), so they are impossible for a product of numbers all $\geq 1$. Choice (C) $\tfrac{7}{2} = 3.5$ matches the first three factors only. Choice (E) $8$ would require an extra factor (a seventh one, $1 + \tfrac{1}{7}$). Only (D) $7$ is consistent — exactly the last numerator after telescoping.
CCSS standards used (min grade 5)
4.NF.B.3Understand a fraction with numerator greater than one as sum of unit fractions (Rewriting each $1 + \tfrac{1}{k}$ as the single fraction $\tfrac{k+1}{k}$ and reading $\tfrac{7}{1}$ as $7$.)4.OA.C.5Generate a number or shape pattern following a given rule (Using small cases ($n = 2, 3, 4$ factors) to spot the rule "the product equals the last numerator" before applying it to all six factors.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying $\tfrac{2}{1}\cdot\tfrac{3}{2}\cdot\tfrac{4}{3}\cdot\tfrac{5}{4}\cdot\tfrac{6}{5}\cdot\tfrac{7}{6}$ and cancelling common factors top-and-bottom (the telescoping step).)
⭐ This AMC 8 problem only needs Grade 5 fraction multiplication you already know — once you spot the cancellation pattern, the answer falls out in one line!
⭐ This AMC 8 problem only needs Grade 5 fraction multiplication you already know — once you spot the cancellation pattern, the answer falls out in one line!
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