Competition · AMC preparation · step 4 of 4

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 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

Count all tile pairs

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

Set up the square test

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

Grade 8 "use square roots" tells us a product is a square exactly when the two halves share the same non-square part.

The product of the tile number and the die number is a perfect square exactly when the two numbers share the same square-free part — the same leftover once each side's largest perfect-square factor is taken out.

▸ Why?

If both numbers hide the same leftover s, write them as s · k² and s · m²; their factors then regroup into (s · k · m) · (s · k · m), a whole number times itself.

▸ Why?

The factors s · k² · s · m² may be reordered to bring the two s's, the two k's, and the two m's next to each other.

Commutativity of multiplicationa rows of b and b rows of a count the same array of dots — rows and columns just swap.

▸ Why?

Once they are side by side, the factors may be regrouped as (s · k · m) · (s · k · m) without changing the product.

Associativity of multiplicationMultiplying three numbers, grouping any two first gives the same product.

▸ Why?

A whole number is a perfect square exactly when every prime factor appears an even number of times, and the product reaches that state only when the two numbers carry the same odd-appearing primes — the same square-free part.

▸ Why?

Each whole number breaks into primes in only one way, so "how many times a prime appears" is a fixed count: squaring a number doubles every one of those counts (all even), and any number whose counts are all even can be halved back into a whole-number root — that is precisely what being a perfect square means.

Unique prime factorizationEvery whole number above 1 is a product of primes in exactly one way (apart from order).

▸ Why?

Multiplying the two numbers pools their prime factors, so each prime's total count is its count in the first number plus its count in the second — and that sum is even only when both counts share the same parity.

Associativity of multiplicationMultiplying three numbers, grouping any two first gives the same product.

8.EE.A.2Identify Subproblems

3STEP 3

Check each tile in turn

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

Divide favorable by total

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

P = 11/60 → (C)

Answer

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 $\frac{11}{60}$ .

Count all tile pairs
Set up the square test
Check each tile in turn
Divide favorable by total

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Count all tile pairs

Set up the square test

Check each tile in turn

Divide favorable by total

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