Competition · AMC preparation · step 4 of 4

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

Count all outcomes

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

Total outcomes = 4 × 4 = 16

2STEP 2

Find the possible range

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

51 ≤ N ≤ 84

3STEP 3

List nearby perfect squares

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

Check which squares are reachable

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

Form and reduce the probability

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

P(N is a perfect square) = 2/16 = 1/8 → (B)

Forming a probability as (favorable outcomes)/(all outcomes) for equally likely outcomes is the Grade 7 probability-model recipe.

Since the 16 possible spinner pairs are equally likely and exactly two of them make N a perfect square, the probability that N is a perfect square is 2 of those 16 pairs.

▸ Why?

Each spinner has 4 equal regions and the two spins do not affect each other, so pairing every one of the 4 Spinner-A results with every one of the 4 Spinner-B results makes 4 equal groups of 4 pairs, which is 16 pairs in all.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

N is built as ten times Spinner A plus Spinner B, so Spinner A is the tens digit and Spinner B is the ones digit, which pins every N between 51 and 84.

Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

▸ Why?

Between 51 and 84 the only perfect squares are 64 and 81, and each splits into a tens digit and a ones digit the spinners can actually land on.

▸ Why?

A perfect square is a whole number times itself, and checking 7 times 7 = 49, 8 times 8 = 64, 9 times 9 = 81, and 10 times 10 = 100 shows only 64 and 81 fall between 51 and 84.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

Reading 64 as 6 tens and 4 ones, and 81 as 8 tens and 1 one, shows their digits sit on the spinners: 6 and 8 on Spinner A, and 4 and 1 on Spinner B, so both come from a real pair.

Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

▸ Why?

When every one of the 16 pairs is equally likely, the probability of an event is just how many pairs favor it out of all 16 pairs, so the two favorable pairs give 2 out of 16.

Probability of equally-likely outcomesWhen every outcome is equally likely, an event's probability is its count of favorable outcomes divided by the total number of outcomes.

7.SP.C.7Make A Systematic List

Answer

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!

Count all outcomes
Find the possible range
List nearby perfect squares
Check which squares are reachable
Form and reduce the probability

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Count all outcomes

Find the possible range

List nearby perfect squares

Check which squares are reachable

Form and reduce the probability

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