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

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Problem

Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 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: 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 $n$.

Givens: Ana and Bonita have a fixed age gap of $n$ years; Last year: Ana's age was $5 \times$ Bonita's age; This year: Ana's age is Bonita's age squared; Answer choices: (A) $3$, (B) $5$, (C) $9$, (D) $12$, (E) $15$

Unknowns: The age gap $n$

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities

Tool #6 (Guess and Check): Ana's age this year is a perfect square, so try small squares ($1, 4, 9, 16, 25, \ldots$) 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.

Execute — Answer: D

#6 Guess and Check 3.OA.C.7 Step 1

List the small perfect squares — these are the possible values for Ana's age this year.
Square roots give Bonita's matching age: $A = 1 \Rightarrow B = 1$, $A = 4 \Rightarrow B = 2$, $A = 9 \Rightarrow B = 3$, $A = 16 \Rightarrow B = 4$, $A = 25 \Rightarrow B = 5$.

$A = B^2$ candidates: $(A, B) = (1, 1), (4, 2), (9, 3), (16, 4), (25, 5), \ldots$

💡 Ana's age is a perfect square — so try the squares in order.

#6 Guess and Check 3.OA.A.3 Step 2

Check each candidate against last year's rule, $A - 1 = 5(B - 1)$.
For $(16, 4)$: $A - 1 = 15$ and $5(B - 1) = 5 \cdot 3 = 15$.
They match.
(The other candidates fail: $(1, 1)$ gives $0 = 0$ but Bonita would have been $0$ — not yet born; $(4, 2)$ gives $3 \ne 5$; $(9, 3)$ gives $8 \ne 10$; $(25, 5)$ gives $24 \ne 20$.)

$$(A, B) = (16, 4):\ A - 1 = 15,\ 5(B - 1) = 5 \cdot 3 = 15\ \checkmark$$

💡 Plug each pair into 'last year' and stop at the one that fits.

#6 Guess and Check 1.OA.A.1 Step 3

With Ana $= 16$ and Bonita $= 4$, the age gap is $n = 16 - 4 = 12$.
This matches choice (D).

$$n = A - B = 16 - 4 = 12 \;\Rightarrow\; \textbf{(D)}$$

💡 Subtract Bonita's age from Ana's to get the constant gap.

#3 Eliminate Possibilities 4.OA.B.4 Step 4

Sanity-eliminate using a divisibility shortcut.
Last year's rule $A - 1 = 5(B - 1)$ rearranges to $A - B = 4(B - 1)$, so the gap $n = A - B$ must be a multiple of $4$.
Among the choices, only $12$ is a multiple of $4$ — (A) $3$, (B) $5$, (C) $9$, (E) $15$ all leave remainders.

$A - 1 = 5(B - 1) \;\Rightarrow\; A - B = 4(B - 1) \;\Rightarrow\; 4 \mid n;$ only $12 = 4 \cdot 3$ qualifies $\;\Rightarrow\; \textbf{(D)}$

💡 Even without finding the ages, the divisibility-by-$4$ filter picks out (D) immediately.

[1] #6 3.OA.C.7 List the small perfect squares — these are the possible values for Ana's age thi

[2] #6 3.OA.A.3 Check each candidate against last year's rule, $A - 1 = 5(B - 1)$. For $(16, 4)$

[3] #6 1.OA.A.1 With Ana $= 16$ and Bonita $= 4$, the age gap is $n = 16 - 4 = 12$. This matches

[4] #3 4.OA.B.4 Sanity-eliminate using a divisibility shortcut. Last year's rule $A - 1 = 5(B -

Review

Reasonableness: Check both clues with Ana $= 16$, Bonita $= 4$. Last year: Ana was $15$, Bonita was $3$, and $15 = 5 \cdot 3$. This year: Ana is $16 = 4^2$, matching Bonita's age squared. The gap stays $12$ every year. Everything fits, so $n = 12$ is correct.

Alternative: Tool #13 (Algebra): set $A = B^2$ and $A - 1 = 5(B - 1)$, substitute to get $B^2 - 1 = 5B - 5$, factor as $(B - 1)(B + 1) = 5(B - 1)$, divide by $B - 1$ (since $B \ne 1$) to get $B + 1 = 5$, so $B = 4$, $A = 16$, $n = 12$.

CCSS standards used (min grade 4)

1.OA.A.1 Solve addition and subtraction word problems within 20 (Computing the age gap $n = A - B = 16 - 4 = 12$.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Checking the 'last year, $5$ times as old' rule by multiplying $5 \cdot 3 = 15$.)
3.OA.C.7 Fluently multiply and divide within 100 (Listing the perfect squares $1, 4, 9, 16, 25$ to test as Ana's age.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that $A - B = 4(B - 1)$ forces $n$ to be a multiple of $4$, eliminating four of the five choices.)

⭐ 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 $12$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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