AMC 8 · 2008 · #24

Grade 8 counting

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

probability-basicperfect-squaressystematic-enumerationfraction-arithmetic caseworksystematic-enumeration ↑ Prerequisites: probability-basicperfect-squares

📏 Short solution 💡 3 insights

Problem

Ten tiles labeled 1 through 10 lie face down. We turn one tile up at random and roll a standard die. What is the probability that (tile number) × (die number) is a perfect square?

Pick an answer.

(A)

$\frac{1}{10}$

(B)

$\frac{1}{6}$

(C)

$\frac{11}{60}$

(D)

$\frac{1}{5}$

(E)

$\frac{7}{30}$

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Counting Carefully

Probability with a finite, equally likely sample space is $\frac{favorable}{total}$ , so the work is pure counting — Tool #13. The total 10 × 6 = 60 is easy. For the favorable count, we use Tool #7 and split the job by tile: fix T, ask which D ∈ {1, …, 6} makes T · D a square. The square-free part of T tells us exactly which D works, so each subproblem is a one-line check.

1STEP 1

The tile gives 10 choices, the die gives 6, and they are independent, so there are 10 × 6 = 60 equally likely pairs.

Total pairs = 10 × 6 = 60

2STEP 2

Write T = s · k² with s square-free; then T · D is a square exactly when D = s · m² shares the same square-free part s.

T = s · k², D = s · m² ⟹ T · D = s² · (km)² = (skm)²

3STEP 3

Go tile by tile: find each T's square-free part s, then list every die D ∈ {1, …, 6} of the form s · m².

T & s & valid D ; 1 & 1 & 1, 4 ; 2 & 2 & 2 ; 3 & 3 & 3 ; 4 & 1 & 1, 4 ; 5 & 5 & 5 ; 6 & 6 & 6 ; 7 & 7 & none ; 8 & 2 & 2 ; 9 & 1 & 1, 4 ; 10 & 10 & none

4STEP 4

Collecting the winners gives 11 favorable (T, D) pairs, so the probability is $\frac{11}{60}$ .

P =

\frac{11}{60}

→ (C)

Answer

\frac{11}{60}

Spot-check a few pairs: 1 · 4 = 4 = 2² (good), 8 · 2 = 16 = 4² (good), 9 · 4 = 36 = 6² (good). For tiles 7 and 10, the square-free parts (7 and 10) both exceed 6, so no die can fix them — that matches the empty rows. The total 11 is close to

\frac{1}{6}

of 60 = 10, so

\frac{11}{60}

≈ 0.183 sits between answers (B)

\frac{1}{6}

≈ 0.167 and (D)

\frac{1}{5}

= 0.20 — only choice (C) matches an exact count of 11.

💡Key takeaway

List, don't guess. Take each tile in turn, ask which die value squares the product, and count the winners. Eleven hits out of sixty gives $\frac{11}{60}$ .

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