AMC 8 · 2018 · #2

Grade 5 arithmeticpattern
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
fraction-arithmeticfraction-multiplicationpattern-recognition pattern-recognitionidentify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic
📏 Short solution 💡 2 insights
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Problem
Compute the value of the product (1+11\frac{1}{1})(1+12\frac{1}{2})(1+13\frac{1}{3})(1+14\frac{1}{4})(1+15\frac{1}{5})(1+16\frac{1}{6}) and match the result to one of the five answer choices.

Pick an answer.

(A)
$\frac{7}{6}$
(B)
$\frac{4}{3}$
(C)
$\frac{7}{2}$
(D)
7
(E)
8

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Each factor 1 + 1k\frac{1}{k} rewrites cleanly as k+1k\frac{k+1}{k}, so the product becomes 21\frac{2}{1}·32\frac{3}{2}·43\frac{4}{3}…76\frac{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.

1STEP 1

Rewrite each factor 1 + 1k\frac{1}{k} as k+1k\frac{k+1}{k}, so the six factors become 21\frac{2}{1}, 32\frac{3}{2}, 43\frac{4}{3}, 54\frac{5}{4}, 65\frac{6}{5}, 76\frac{7}{6}.

1+1k\frac{1}{k} = k+1k\frac{k+1}{k} → 21\frac{2}{1}, 32\frac{3}{2}, 43\frac{4}{3}, 54\frac{5}{4}, 65\frac{6}{5}, 76\frac{7}{6}
2STEP 2

Test smaller cases (Tool #9): 2 factors give 3, 3 give 4, 4 give 5 — the pattern leaves only the last numerator.

21\frac{2}{1}·32\frac{3}{2} = 3, 21\frac{2}{1}·32\frac{3}{2}·43\frac{4}{3} = 4, 21\frac{2}{1}·32\frac{3}{2}·43\frac{4}{3}·54\frac{5}{4} = 5
3STEP 3

Apply the telescoping cancellation to all six factors: every inner number cancels, leaving 21\frac{2}{1}·32\frac{3}{2}·43\frac{4}{3}·54\frac{5}{4}·65\frac{6}{5}·76\frac{7}{6} = 71\frac{7}{1}.

21\frac{2}{1}·32\frac{3}{2}·43\frac{4}{3}·54\frac{5}{4}·65\frac{6}{5}·76\frac{7}{6} = 71\frac{7}{1}
4STEP 4

Read off the value: 71\frac{7}{1} = 7, which matches choice (D).

71\frac{7}{1} = 7 → (D)
Answer
7
Each factor is greater than 1, so the product must be greater than 1 — that already eliminates nothing, but the smallest factor is 76\frac{7}{6} ≈ 1.17 and the largest is 2, so a rough estimate 2 × 1.5 × 1.33 × 1.25 × 1.2 × 1.17 ≈ 7 matches the exact answer. The pattern check is even stronger: with n factors of the form 1+1k\frac{1}{k} for k=1… n, the product equals n+1. Here n = 6, so the answer is 7 — choice (D).
💡Key takeaway

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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