AMC 10 · 2019 · #3

Grade 4 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

linear-equations-two-varsystems-of-equationsdifference-of-squaresperfect-squares convert-to-algebraguess-and-check ↑ Prerequisites: linear-equations-two-varsystems-of-equations

📏 Short solution 💡 2 insights

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Problem

Ana and Bonita were born on the same calendar date but in different years, with Ana being n years older. Last year Ana's age was 5 times Bonita's age. This year Ana's age is the square of Bonita's age. Find the age gap n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

Tool #6 (Guess and Check): Ana's age this year is a perfect square, so try small squares (1, 4, 9, 16, 25, …) for Ana and read off Bonita as the square root. For each candidate, check whether last year Ana was exactly 5 times Bonita. Tool #5 (Pattern): the gap n = A - B is forced once we know A and B. Tool #3 (Eliminate): last year A - 1 = 5(B - 1) means A - B = 4(B - 1), so n must be a multiple of 4. Only choice (D) 12 is a multiple of 4 — instant answer.

1STEP 1

List the small perfect squares as Ana's possible ages this year; each square root is Bonita's matching age.

A = B² candidates: (A, B) = (1, 1), (4, 2), (9, 3), (16, 4), (25, 5), …

2STEP 2

Test each pair against last year's rule A - 1 = 5(B - 1); only (16, 4) fits, since 15 = 5 · 3.

(A, B) = (16, 4): A - 1 = 15, 5(B - 1) = 5 · 3 = 15 ✓

3STEP 3

Ana 16 minus Bonita 4 gives the constant age gap n = 16 - 4 = 12, choice (D).

n = A - B = 16 - 4 = 12 → (D)

4STEP 4

Rearranged, A - B = 4(B - 1), so the gap must be a multiple of 4 — only 12 qualifies, giving (D).

A - 1 = 5(B - 1) → A - B = 4(B - 1) → 4 ∣ n; only 12 = 4 · 3 qualifies → (D)

Answer

Check both clues with Ana = 16, Bonita = 4. Last year: Ana was 15, Bonita was 3, and 15 = 5 · 3. This year: Ana is 16 = 4², matching Bonita's age squared. The gap stays 12 every year. Everything fits, so n = 12 is correct.

💡Key takeaway

This AMC 10 problem only needs Grade 4 "factors and multiples" you already know — Ana's age must be a perfect square and the gap must be a multiple of 4, so try 16 and 4 and the gap is 12.

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