AMC 10 · 2022 · #5

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

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

📏 Medium solution 💡 2 insights

Problem

Evaluate ((1 +

\frac{1}{3}

)(1 +

\frac{1}{5}

)(1 +

\frac{1}{7}

))/(√((1 -

\frac{1}{3²}

)(1 -

\frac{1}{5²}

)(1 -

\frac{1}{7²}

))). The numerator and the expression under the square root share a hidden link: each 1 -

\frac{1}{n²}

factors as (1 -

\frac{1}{n}

)(1 +

\frac{1}{n}

), so the same (1 +

\frac{1}{n}

) factors appear in both places.

Pick an answer.

(A)

$\sqrt3$

(B)

(C)

$\sqrt{15}$

(D)

(E)

$\sqrt{105}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 a² - b² = (a - b)(a + b) applied to 1 - $\frac{1}{n²}$ . 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 √(3), 2, √(15), 4, √(105), so once we land on a value we can match it.

1STEP 1

Apply the difference of squares to each factor under the radical: 1 - $\frac{1}{n²}$ = (1 - $\frac{1}{n}$ )(1 + $\frac{1}{n}$ ) for n = 3, 5, 7.

D = (1 -

\frac{1}{3}

)(1 +

\frac{1}{3}

)(1 -

\frac{1}{5}

)(1 +

\frac{1}{5}

)(1 -

\frac{1}{7}

)(1 +

\frac{1}{7}

)

2STEP 2

With M = (1 - $\frac{1}{3}$ )(1 - $\frac{1}{5}$ )(1 - $\frac{1}{7}$ ), the numerator N reappears inside the root as N · M, so the whole thing collapses to √( $\frac{N}{M}$ ).

\frac{N}{√(D)}

\frac{N}{(√(N · M))}

= √(

\frac{N}{M}

)

3STEP 3

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

N =

\frac{4}{3}

\frac{6}{5}

\frac{8}{7}

, M =

\frac{2}{3}

\frac{4}{5}

\frac{6}{7}

4STEP 4

Form $\frac{N}{M}$ — the 3 · 5 · 7 denominators cancel, then 4 and 6 cancel too, leaving $\frac{N}{M}$ = 4.

\frac{N}{M}

= (

\frac{4 · 6 · 8}{3 · 5 · 7}

)/(

\frac{2 · 4 · 6}{3 · 5 · 7}

) =

\frac{4 · 6 · 8}{2 · 4 · 6}

\frac{8}{2}

= 4

5STEP 5

Take the square root of the ratio: √4 gives the final value 2.

√(

\frac{N}{M}

) = √(4) = 2 → (B)

Answer

Plug a quick sanity check. Each 1 +

\frac{1}{n}

is slightly more than 1, and each 1 -

\frac{1}{n}

is slightly less than 1, so

\frac{1 + \frac{1}{n}}{1 - \frac{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 · 6 · 8 = 2 · 4 · 6 · 4 make the final √(4) = 2 exact, not just approximate.

💡Key takeaway

This AMC 10 problem only needs Grade 8 "1 - $\frac{1}{n²}$ = (1 - $\frac{1}{n}$ )(1 + $\frac{1}{n}$ )" — once that factoring is in place, the big fraction collapses to √(4) = 2.

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