AMC 8 · 2019 · #17

Easy mode Grade 5

Problem

Below is a long product of fractions. Look closely at the pattern.

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

Each fraction follows the same shape. The numerator is two numbers multiplied together. The denominator is also two numbers multiplied together.

To see the pattern, try the first fraction. The top is $1 \cdot 3$ , and the bottom is $2 \cdot 2$ . The next fraction shifts everything up by one: top is $2 \cdot 4$ , bottom is $3 \cdot 3$ . The last fraction has top $98 \cdot 100$ , bottom $99 \cdot 99$ .

Multiply all of these fractions together. What is the value of the whole product?

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.

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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 just two factors

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

Redo the small product cleanly

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

Read off the pattern

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

Split and telescope

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 chains

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!

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Try just two factors

Redo the small product cleanly

Read off the pattern

Split and telescope

Combine the two chains

Review

CCSS standards used (min grade 5)

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