AMC 10 · 2022 · #5

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition

fraction-arithmeticdifference-of-squarespolynomial-factoringexponents identify-subproblemspattern-recognition ↑ Prerequisites: fraction-arithmeticdifference-of-squares

📏 Medium solution 💡 2 insights

Problem

What is the value of $\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?$

Pick an answer.

(A)

$\sqrt3$

(B)

(C)

$\sqrt{15}$

(D)

(E)

$\sqrt{105}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Evaluate $\dfrac{\left(1 + \tfrac{1}{3}\right)\left(1 + \tfrac{1}{5}\right)\left(1 + \tfrac{1}{7}\right)}{\sqrt{\left(1 - \tfrac{1}{3^2}\right)\left(1 - \tfrac{1}{5^2}\right)\left(1 - \tfrac{1}{7^2}\right)}}$. The numerator and the expression under the square root share a hidden link: each $1 - \tfrac{1}{n^2}$ factors as $\left(1 - \tfrac{1}{n}\right)\left(1 + \tfrac{1}{n}\right)$, so the same $\left(1 + \tfrac{1}{n}\right)$ factors appear in both places.

Givens: Numerator: $N = \left(1 + \tfrac{1}{3}\right)\left(1 + \tfrac{1}{5}\right)\left(1 + \tfrac{1}{7}\right)$; Denominator: $\sqrt{D}$ where $D = \left(1 - \tfrac{1}{3^2}\right)\left(1 - \tfrac{1}{5^2}\right)\left(1 - \tfrac{1}{7^2}\right)$; Identity: $1 - \tfrac{1}{n^2} = \left(1 - \tfrac{1}{n}\right)\left(1 + \tfrac{1}{n}\right)$ (difference of squares); Answer choices: (A) $\sqrt{3}$, (B) $2$, (C) $\sqrt{15}$, (D) $4$, (E) $\sqrt{105}$

Unknowns: The simplified numerical value of $\dfrac{N}{\sqrt{D}}$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #13 Convert to Algebra, #6 Guess and Check

Tool #7 (Subproblems) breaks the expression into three pieces: (a) factor the inside of the square root, (b) simplify the resulting ratio, (c) compute the final arithmetic. Tool #13 (Convert to Algebra) supplies the lever for step (a) — the difference of squares $a^2 - b^2 = (a - b)(a + b)$ applied to $1 - \tfrac{1}{n^2}$. Once everything is rewritten as a single square root of a clean ratio, the cancellation is immediate. Tool #6 (Guess and Check) acts as the multiple-choice safety net: the answer is one of $\sqrt{3}, 2, \sqrt{15}, 4, \sqrt{105}$, so once we land on a value we can match it.

Execute — Answer: B

#13 Convert to Algebra 7.EE.A.1 Step 1

Apply the difference of squares to each factor under the radical: $1 - \tfrac{1}{n^2} = \left(1 - \tfrac{1}{n}\right)\left(1 + \tfrac{1}{n}\right)$ for $n = 3, 5, 7$.

$$D = \left(1 - \tfrac{1}{3}\right)\left(1 + \tfrac{1}{3}\right)\left(1 - \tfrac{1}{5}\right)\left(1 + \tfrac{1}{5}\right)\left(1 - \tfrac{1}{7}\right)\left(1 + \tfrac{1}{7}\right)$$

💡 Difference of squares is Grade 7 "factor and expand linear expressions" — splits each $1 - \tfrac{1}{n^2}$ into the two pieces we already need.

#7 Identify Subproblems 8.EE.A.2 Step 2

Group the factors of $D$ so the $(1 + \tfrac{1}{n})$ pieces line up with the numerator $N$.
Let $M = \left(1 - \tfrac{1}{3}\right)\left(1 - \tfrac{1}{5}\right)\left(1 - \tfrac{1}{7}\right)$.
Then $D = N \cdot M$, so $\sqrt{D} = \sqrt{N \cdot M} = \sqrt{N} \cdot \sqrt{M}$.
The full expression becomes $\dfrac{N}{\sqrt{N} \cdot \sqrt{M}} = \dfrac{\sqrt{N}}{\sqrt{M}} = \sqrt{\dfrac{N}{M}}$.

$$\dfrac{N}{\sqrt{D}} = \dfrac{N}{\sqrt{N \cdot M}} = \sqrt{\dfrac{N}{M}}$$

💡 $\tfrac{a}{\sqrt{a}} = \sqrt{a}$ collapses the heavy fraction into a clean square root — Grade 8 "use square root symbols to represent solutions".

#7 Identify Subproblems 5.NF.A.1 Step 3

Compute the simple fractions in $N$ and $M$ (common-denominator addition/subtraction).

$$N = \tfrac{4}{3} \cdot \tfrac{6}{5} \cdot \tfrac{8}{7}, \quad M = \tfrac{2}{3} \cdot \tfrac{4}{5} \cdot \tfrac{6}{7}$$

💡 Each $1 \pm \tfrac{1}{n} = \tfrac{n \pm 1}{n}$ — Grade 5 "add and subtract fractions with unlike denominators" (here just unit fractions on top of $1$).

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

Form the ratio $\tfrac{N}{M}$.
Denominators $3 \cdot 5 \cdot 7$ cancel; in the remaining numerators, the $4$ and $6$ cancel too.

$$\dfrac{N}{M} = \dfrac{\tfrac{4 \cdot 6 \cdot 8}{3 \cdot 5 \cdot 7}}{\tfrac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}} = \dfrac{4 \cdot 6 \cdot 8}{2 \cdot 4 \cdot 6} = \dfrac{8}{2} = 4$$

💡 Multiplying and cancelling fractions — Grade 5 "multiply a fraction by a fraction". Every factor lines up cleanly.

#6 Guess and Check 8.EE.A.2 Step 5

Take the square root.

$$\sqrt{\dfrac{N}{M}} = \sqrt{4} = 2 \;\Rightarrow\; \textbf{(B)}$$

💡 $\sqrt{4} = 2$ — Grade 8 "square root symbols". Match against the choices: $2$ is exactly (B).

[1] #13 7.EE.A.1 Apply the difference of squares to each factor under the radical: $1 - \tfrac{1}

[2] #7 8.EE.A.2 Group the factors of $D$ so the $(1 + \tfrac{1}{n})$ pieces line up with the num

[3] #7 5.NF.A.1 Compute the simple fractions in $N$ and $M$ (common-denominator addition/subtrac

[4] #7 5.NF.B.4 Form the ratio $\tfrac{N}{M}$. Denominators $3 \cdot 5 \cdot 7$ cancel; in the r

[5] #6 8.EE.A.2 Take the square root.

Review

Reasonableness: Plug a quick sanity check. Each $1 + \tfrac{1}{n}$ is slightly more than $1$, and each $1 - \tfrac{1}{n}$ is slightly less than $1$, so $\tfrac{1 + 1/n}{1 - 1/n}$ is a bit more than $1$. The product over $n = 3, 5, 7$ should sit comfortably above $1$ but well below $10$, and the square root pulls it even closer to $1$. A value of $2$ is exactly the right magnitude — and the cancellations $4 \cdot 6 \cdot 8 = 2 \cdot 4 \cdot 6 \cdot 4$ make the final $\sqrt{4} = 2$ exact, not just approximate.

Alternative: Tool #5 (Look for a Pattern). Try a smaller version with just $n = 3$: $\dfrac{1 + \tfrac{1}{3}}{\sqrt{1 - \tfrac{1}{9}}} = \dfrac{4/3}{\sqrt{8/9}} = \dfrac{4/3}{(2\sqrt{2})/3} = \dfrac{4}{2\sqrt{2}} = \sqrt{2}$. So $\dfrac{1 + 1/n}{\sqrt{1 - 1/n^2}} = \sqrt{\dfrac{1 + 1/n}{1 - 1/n}} = \sqrt{\dfrac{n+1}{n-1}}$. The product for $n = 3, 5, 7$ is $\sqrt{\tfrac{4}{2} \cdot \tfrac{6}{4} \cdot \tfrac{8}{6}} = \sqrt{\tfrac{8}{2}} = \sqrt{4} = 2$. Same answer.

CCSS standards used (min grade 8)

5.NF.A.1 Add and subtract fractions with unlike denominators (Computing $1 \pm \tfrac{1}{n} = \tfrac{n \pm 1}{n}$ for $n = 3, 5, 7$.)
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying the simple fractions in $N$ and $M$ and cancelling common factors in the ratio $N/M$.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Factoring $1 - \tfrac{1}{n^2}$ as $\left(1 - \tfrac{1}{n}\right)\left(1 + \tfrac{1}{n}\right)$ — the difference of squares lever.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Collapsing $\tfrac{N}{\sqrt{N \cdot M}}$ to $\sqrt{N/M}$ and finishing with $\sqrt{4} = 2$.)

⭐ This AMC 10 problem only needs Grade 8 "$1 - \tfrac{1}{n^2} = (1 - \tfrac{1}{n})(1 + \tfrac{1}{n})$" — once that factoring is in place, the big fraction collapses to $\sqrt{4} = 2$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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