AMC 10 · 2019 · #3
Easy mode Grade 4Problem
Ana and Bonita have the same birthday, but they were born in different years. Ana is older. The gap between their ages is n years, and that gap never changes.
Last year, Ana's age was exactly 5 times Bonita's age.
This year, Ana's age equals Bonita's age multiplied by itself. (For example, if Bonita is 4 this year, then Ana is 4×4=16.)
What is n, the number of years between their ages?
Pick an answer.
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.
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
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) $15$
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
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: $A = 1 \Rightarrow B = 1$, $A = 4 \Rightarrow B = 2$, $A = 9 \Rightarrow B = 3$, $A = 16 \Rightarrow B = 4$, $A = 25 \Rightarrow B = 5$.
💡 Ana's age is a perfect square — so try the squares in order.
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 $24 \ne 20$.)
💡 Plug each pair into 'last year' and stop at the one that fits.
1.OA.A.1 Step 3 - With Ana $= 16$ and Bonita $= 4$, the age gap is $n = 16 - 4 = 12$.
- This matches choice (D).
💡 Subtract Bonita's age from Ana's to get the constant gap.
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.
💡 Even without finding the ages, the divisibility-by-$4$ filter picks out (D) immediately.
3.OA.C.7 List the small perfect squares — these are the possible values for Ana's age thi 3.OA.A.3 Check each candidate against last year's rule, $A - 1 = 5(B - 1)$. For $(16, 4)$ 1.OA.A.1 With Ana $= 16$ and Bonita $= 4$, the age gap is $n = 16 - 4 = 12$. This matches 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.1Solve addition and subtraction word problems within 20 (Computing the age gap $n = A - B = 16 - 4 = 12$.)3.OA.A.3Solve 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.7Fluently multiply and divide within 100 (Listing the perfect squares $1, 4, 9, 16, 25$ to test as Ana's age.)4.OA.B.4Find 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 $12$.
⭐ 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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