AMC 8 · 2022 · #12

Grade 7 probabilitycounting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

probability-basicperfect-squaressystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: probability-basicperfect-squares

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Spinner A is split into 4 equal regions labeled 5, 6, 7, 8, and Spinner B is split into 4 equal regions labeled 1, 2, 3, 4. Spin both, then form the two-digit number N = 10 × (Spinner A) + (Spinner B). Find the probability that N is a perfect square.

Pick an answer.

(A)

$\dfrac{1}{16}$

(B)

$\dfrac{1}{8}$

(C)

$\dfrac{1}{4}$

(D)

$\dfrac{3}{8}$

(E)

$\dfrac{1}{2}$

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The sample space has only 4 × 4 = 16 outcomes — small enough to enumerate exhaustively, so Tool #2 (Systematic List) is the perfect fit. Tool #7 (Identify Subproblems) splits the work into two clean pieces: (a) count the total outcomes (16), and (b) count how many produce a perfect square. For (b), instead of squaring every two-digit value, use Tool #6 (Guess and Check) on the small set of perfect squares actually in range — 8² = 64 and 9² = 81 are the only candidates because 7² = 49 ＜ 51 and 10² = 100 ＞ 84. Then check whether those squares' digits really match what the spinners can produce.

1STEP 1

Two independent spinners with 4 results each give 4 × 4 = 16 equally likely ordered pairs (A, B).

Total outcomes = 4 × 4 = 16

2STEP 2

Spinner A gives the tens digit and Spinner B the ones, so N lies in 51 ≤ N ≤ 84.

51 ≤ N ≤ 84

3STEP 3

Squares near the range: 7² = 49 is too small and 10² = 100 too big, so only 64 (= 8²) and 81 (= 9²) fit.

7² = 49, 8² = 64, 9² = 81, 10² = 100 → {64, 81}

4STEP 4

N = 64 needs (A, B) = (6, 4) and N = 81 needs (8, 1) — both on the spinners, so there are 2 favorable outcomes.

Favorable pairs = {(6,4), (8,1)} → 2 outcomes

5STEP 5

Probability is favorable over total: $\frac{2}{16}$ , which reduces to $\frac{1}{8}$ — choice (B).

P(N is a perfect square) =

\frac{2}{16}

\frac{1}{8}

→ (B)

Answer

\frac{1}{8}

Out of 16 tiny equally likely outcomes, only 2 are perfect squares — that is a small fraction, so the answer should be small.

\frac{1}{8}

= 0.125 is small and matches choice (B). Choices (D)

\frac{3}{8}

and (E)

\frac{1}{2}

would require 6 or 8 perfect squares among the 16 values, but the range 51–84 obviously cannot contain that many squares (squares get sparser as numbers grow). And

\frac{1}{16}

(only 1 favorable) would miss either 64 or 81.

💡Key takeaway

This AMC 8 problem only needs the Grade 7 "favorable over total" probability model you already know — the perfect-square hunt is just Grade 3 multiplication facts!

More like this