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 numbered $1$ through $10$ are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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) $\times$ (die number) is a perfect square?

Givens: Tiles are numbered $1, 2, 3, \ldots, 10$ (so $10$ equally likely tile outcomes); A standard die shows $1, 2, 3, 4, 5, 6$ (so $6$ equally likely die outcomes); The two draws are independent; Answer choices: (A) $\tfrac{1}{10}$, (B) $\tfrac{1}{6}$, (C) $\tfrac{11}{60}$, (D) $\tfrac{1}{5}$, (E) $\tfrac{7}{30}$

Unknowns: The probability that the product is a perfect square

Understand

Plan

Primary tool: #13 Counting Carefully

Secondary: #7 Break Into Subproblems

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

Execute — Answer: C

#13 Counting Carefully 7.SP.C.8 Step 1

Count the total number of equally likely outcomes.
The tile gives $10$ choices and the die gives $6$ choices, and the two are independent.

$$\text{Total pairs} = 10 \times 6 = 60$$

💡 Grade 7 says compound probabilities use the size of the sample space; here that size is $60$.

#7 Break Into Subproblems 8.EE.A.2 Step 2

Set up the square test.
Write $T = s \cdot k^2$ where $s$ is the square-free part (no repeated prime factor).
Then $T \cdot D$ is a square iff $D$ has the same square-free part $s$, i.e.
$D = s \cdot m^2$ for some integer $m \ge 1$.
So for each tile we just need to find the multiples of $s$ in $\{1, \ldots, 6\}$ whose quotient is a square.

$$T = s \cdot k^2, \; D = s \cdot m^2 \;\Longrightarrow\; T \cdot D = s^2 \cdot (km)^2 = (skm)^2$$

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

#7 Break Into Subproblems 7.SP.C.8 Step 3

Run through the tiles one by one.
For each $T$ find its square-free part $s$, then list every $D \in \{1, \ldots, 6\}$ of the form $s \cdot m^2$.

$$\begin{array}{l|l|l} T & s & \text{valid } D \\\hline 1 & 1 & 1, 4 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \\ 4 & 1 & 1, 4 \\ 5 & 5 & 5 \\ 6 & 6 & 6 \\ 7 & 7 & \text{none} \\ 8 & 2 & 2 \\ 9 & 1 & 1, 4 \\ 10 & 10 & \text{none} \end{array}$$

💡 Each row is a tiny subproblem: "what does the die need to multiply $T$ to a square?" If $s > 6$ no die works.

#13 Counting Carefully 7.SP.C.8 Step 4

Add up the favorable pairs and divide by the total.
The valid $(T, D)$ pairs are $(1,1), (1,4), (2,2), (3,3), (4,1), (4,4), (5,5), (6,6), (8,2), (9,1), (9,4)$ — that is $11$ pairs.

$$P = \dfrac{11}{60} \;\Rightarrow\; \textbf{(C)}$$

💡 Favorable over total: $\tfrac{11}{60}$ does not reduce because $\gcd(11, 60) = 1$.

[1] #13 7.SP.C.8 Count the total number of equally likely outcomes. The tile gives $10$ choices a

[2] #7 8.EE.A.2 Set up the square test. Write $T = s \cdot k^2$ where $s$ is the square-free par

[3] #7 7.SP.C.8 Run through the tiles one by one. For each $T$ find its square-free part $s$, th

[4] #13 7.SP.C.8 Add up the favorable pairs and divide by the total. The valid $(T, D)$ pairs are

Review

Reasonableness: Spot-check a few pairs: $1 \cdot 4 = 4 = 2^2$ (good), $8 \cdot 2 = 16 = 4^2$ (good), $9 \cdot 4 = 36 = 6^2$ (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 $\tfrac{1}{6}$ of $60 = 10$, so $\tfrac{11}{60} \approx 0.183$ sits between answers (B) $\tfrac{1}{6} \approx 0.167$ and (D) $\tfrac{1}{5} = 0.20$ — only choice (C) matches an exact count of $11$.

Alternative: Tool #2 (Make an Organized List): write out all $60$ products in a $10 \times 6$ grid and circle the perfect squares ($1, 4, 9, 16, 25, 36$). You will mark exactly the same $11$ cells — same answer, just slower.

CCSS standards used (min grade 8)

7.SP.C.8 Find probabilities of compound events using organized lists, tables, or simulation (Counting the sample space ($60$ equally likely pairs) and the favorable event ($11$ pairs), then dividing to get $P = \tfrac{11}{60}$.)
8.EE.A.2 Use square root and cube root symbols; know that $\sqrt{2}$ is irrational; evaluate square roots of small perfect squares (Recognising perfect squares via their prime structure: $T \cdot D$ is a square iff $T$ and $D$ share the same square-free part.)

⭐ 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 $\tfrac{11}{60}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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