AMC 8 · 2019 · #17

Grade 5 arithmeticpattern

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

fraction-multiplicationpattern-recognitionsequences-arithmetic pattern-recognitionidentify-subproblems ↑ Prerequisites: fraction-multiplicationpattern-recognition

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

What is the value of the product

$\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?$

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{50}{99}$

(C)

$\frac{9800}{9801}$

(D)

$\frac{100}{99}$

(E)

AMC 8 2019 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: We multiply $98$ fractions of the form $\frac{k \cdot (k+2)}{(k+1) \cdot (k+1)}$, starting at $k = 1$ (so the first factor is $\frac{1 \cdot 3}{2 \cdot 2}$) and ending at $k = 98$ (so the last factor is $\frac{98 \cdot 100}{99 \cdot 99}$). What single number does this big product equal?

Givens: Each factor has the shape $\frac{k(k+2)}{(k+1)^2}$; The first factor uses $k = 1$: $\frac{1 \cdot 3}{2 \cdot 2}$; The last factor uses $k = 98$: $\frac{98 \cdot 100}{99 \cdot 99}$; There are $98$ factors total (one for each $k$ from $1$ to $98$); Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{50}{99}$, (C) $\tfrac{9800}{9801}$, (D) $\tfrac{100}{99}$, (E) $50$

Unknowns: The exact value of the full product as a single fraction

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #7 Identify Subproblems

The product has $98$ factors — way too many to multiply by hand. Tool #9 (Easier Related Problem) says: try the same product with only $2$, then $3$, then $4$ factors first, watch what happens, and conjecture a formula. Tool #5 (Look for a Pattern) reads the small-case results into a clean rule. Tool #7 (Identify Subproblems) gives the cleanest shortcut: split each factor as $\frac{k}{k+1} \cdot \frac{k+2}{k+1}$, which turns the giant product into two telescoping chains that each collapse in one line. Both routes give the same answer; the small-case path is friendlier for a young solver, and the telescoping split confirms it.

Execute — Answer: B

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

Try the easier version with just the first $2$ factors ($k = 1, 2$).
Multiply numerators across and denominators across.

$$\dfrac{1 \cdot 3}{2 \cdot 2} \cdot \dfrac{2 \cdot 4}{3 \cdot 3} = \dfrac{(1 \cdot 3)(2 \cdot 4)}{(2 \cdot 2)(3 \cdot 3)} = \dfrac{24}{36} = \dfrac{2}{3} \cdot \dfrac{1}{2} = \dfrac{1 \cdot 4}{2 \cdot 3} \cdot \dfrac{1}{1}$$

💡 Multiplying a few small fractions is just $\text{(top} \times \text{top}) / (\text{bottom} \times \text{bottom})$ — exactly the Grade 5 fraction-times-fraction skill.

#5 Look for a Pattern 5.NF.B.4 Step 2

Compute the same product cleanly.
With $k = 1, 2$ the product is $\frac{1 \cdot 2 \cdot 3 \cdot 4}{2 \cdot 2 \cdot 3 \cdot 3} = \frac{1 \cdot 4}{2 \cdot 3} = \frac{4}{6} = \frac{2}{3}$.
Now redo it with the first $3$ factors ($k = 1, 2, 3$).

$$2\text{ factors: } \dfrac{1 \cdot 4}{2 \cdot 3} = \dfrac{4}{6}; \quad 3\text{ factors: } \dfrac{1 \cdot 5}{2 \cdot 4} = \dfrac{5}{8}$$

💡 Doing a second small case lets the pattern in the surviving top and bottom numbers start to show.

#5 Look for a Pattern 4.OA.C.5 Step 3

Tabulate the small cases and read off the pattern.
The number of factors $n$ tells us the largest $k$ used.
The result is always $\frac{\text{small number}}{\text{slightly larger number}}$, and the surviving values are exactly the missing 'ends'.

$$n=2: \dfrac{1 \cdot 4}{2 \cdot 3}, \quad n=3: \dfrac{1 \cdot 5}{2 \cdot 4}, \quad n=4: \dfrac{1 \cdot 6}{2 \cdot 5}, \quad \ldots, \quad n: \dfrac{1 \cdot (n+2)}{2 \cdot (n+1)}$$

💡 Generating a number pattern from a given rule (Grade 4) lets us guess the formula from three data points.

#7 Identify Subproblems 5.NF.B.4 Step 4

Confirm the formula by splitting each factor and watching the telescoping.
Write $\frac{k(k+2)}{(k+1)^2} = \frac{k}{k+1} \cdot \frac{k+2}{k+1}$.
The full product becomes two chains: $\left(\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{98}{99}\right) \cdot \left(\frac{3}{2} \cdot \frac{4}{3} \cdots \frac{100}{99}\right)$.
In each chain, every middle number cancels with the same number in the neighboring fraction.

$$\text{Chain 1} = \dfrac{1}{99}, \quad \text{Chain 2} = \dfrac{100}{2} = 50$$

💡 Splitting one hard fraction into two friendlier ones, and noticing equal numbers cancel top-to-bottom, is the heart of Tool #7.

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

Combine the two collapsed chains to get the final value, and check against the formula from the small cases (with $n = 98$, the formula gives $\frac{1 \cdot 100}{2 \cdot 99} = \frac{100}{198} = \frac{50}{99}$).

$$P = \dfrac{1}{99} \cdot 50 = \dfrac{50}{99} \;\Rightarrow\; \textbf{(B)}$$

💡 Both the small-case pattern and the telescoping shortcut land on the same $\frac{50}{99}$, so the answer is locked in.

[1] #9 5.NF.B.4 Try the easier version with just the first $2$ factors ($k = 1, 2$). Multiply nu

[2] #5 5.NF.B.4 Compute the same product cleanly. With $k = 1, 2$ the product is $\frac{1 \cdot

[3] #5 4.OA.C.5 Tabulate the small cases and read off the pattern. The number of factors $n$ tel

[4] #7 5.NF.B.4 Confirm the formula by splitting each factor and watching the telescoping. Write

[5] #5 5.NF.B.4 Combine the two collapsed chains to get the final value, and check against the f

Review

Reasonableness: Each individual factor $\frac{k(k+2)}{(k+1)^2} = 1 - \frac{1}{(k+1)^2}$ is just slightly less than $1$ (e.g., $\frac{1 \cdot 3}{4} = 0.75$, $\frac{2 \cdot 4}{9} \approx 0.889$, $\frac{3 \cdot 5}{16} \approx 0.9375$, climbing toward $1$). Multiplying $98$ numbers each less than $1$ must give something less than $1$, and the early factors drag it well below $1$. The value $\frac{50}{99} \approx 0.505$ — roughly $\frac{1}{2}$ — fits perfectly. Answer (A) $\frac{1}{2}$ is dangerously close but the formula gives exactly $\frac{50}{99}$, not $\frac{1}{2}$ (since $\frac{50}{99} \ne \frac{49.5}{99}$). Answer (C) $\frac{9800}{9801}$ is too close to $1$, (D) and (E) are larger than $1$, so they cannot be right.

Alternative: Tool #5 (Look for a Pattern) alone: from $n = 2, 3, 4$ the products are $\frac{4}{6}, \frac{5}{8}, \frac{6}{10}$ — i.e. $\frac{n+2}{2(n+1)}$. Plugging $n = 98$ gives $\frac{100}{2 \cdot 99} = \frac{50}{99}$ without any telescoping argument at all.

CCSS standards used (min grade 5)

4.OA.C.5 Generate a number or shape pattern following a given rule (Reading the small-case results ($n = 2, 3, 4$) as a pattern and extending the rule to the closed-form expression $\frac{n+2}{2(n+1)}$.)
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying several fractions of the form $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$ in both the small cases and the telescoping split, then collapsing the chains into $\frac{1}{99}$ and $50$.)

⭐ This AMC 8 problem only needs Grade 5 fraction multiplication you already know — try a few small cases, spot the pattern, and the giant product solves itself!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 5)

More like this