AMC 10 · 2009 · #6
Grade 4 arithmeticKiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
Pick an answer.
AMC 10 2009 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: Kiana has two older twin brothers. Because they are twins, the two brothers are the same age, and both are older than Kiana. The three ages multiply to $128$. Find the sum of the three ages.
Givens: The two brothers are twins, so they share one age.; Both brothers are older than Kiana.; The product of all three ages is $128$.; Ages are whole numbers.
Unknowns: The brothers' shared age.; Kiana's age.; The sum of the three ages.
Understand
Restated: Kiana has two older twin brothers. Because they are twins, the two brothers are the same age, and both are older than Kiana. The three ages multiply to $128$. Find the sum of the three ages.
Givens: The two brothers are twins, so they share one age.; Both brothers are older than Kiana.; The product of all three ages is $128$.; Ages are whole numbers.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
Name the twin age $t$ and Kiana's age $k$. Since the twins share an age, the product becomes $t \times t \times k = 128$, so $t$ shows up twice. That lets me list only the whole-number twin ages whose square divides $128$, then throw out the ones that break the 'older than Kiana' rule.
Execute — Answer: D
4.OA.A.3 Step 1 Name the ages
- Let $t$ be the age each twin shares and $k$ be Kiana's age.
- The twins are the same age, so the product of the three ages is $t \times t \times k$.
- Setting this equal to the given product gives one equation.
💡 Using one letter for the twins captures the fact that their two ages are really the same number.
4.OA.B.4 Step 2 List the twin ages that fit
- Because $128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$, the twin age $t$ must be a whole number whose square divides $128$.
- Test $t = 1, 2, 4, 8$ and read off Kiana's matching age $k = 128 \div t^2$: $t=1 \Rightarrow k=128$, $t=2 \Rightarrow k=32$, $t=4 \Rightarrow k=8$, $t=8 \Rightarrow k=2$.
💡 Since the twin age is used twice, only ages whose square fits inside $128$ are even possible.
4.NBT.A.2 Step 3 Keep only older twins
- The brothers must be older than Kiana, so I need $t > k$.
- Check each case: $1 > 128$ is false, $2 > 32$ is false, $4 > 8$ is false, but $8 > 2$ is true.
- Only the twin age $t = 8$ with Kiana's age $k = 2$ survives.
💡 The 'older brothers' fact is a filter that removes every factor set except the one where the twins are bigger than Kiana.
4.NBT.B.4 Step 4 Add the three ages
- The three ages are $8$, $8$, and $2$.
- Their sum is $2t + k = 8 + 8 + 2 = 18$.
- So the sum of the three ages is $18$, which is answer (D).
💡 Once the only valid ages are found, adding them is the last easy step.
4.OA.A.3 Let $t$ be the age each twin shares and $k$ be Kiana's age. The twins are the sa 4.OA.B.4 Because $128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$, the twi 4.NBT.A.2 The brothers must be older than Kiana, so I need $t > k$. Check each case: $1 > 4.NBT.B.4 The three ages are $8$, $8$, and $2$. Their sum is $2t + k = 8 + 8 + 2 = 18$. So Review
Reasonableness: Check the survivor directly: $8 \times 8 \times 2 = 128$ matches the product, and both brothers ($8$) are older than Kiana ($2$), so every condition holds. The sum $8 + 8 + 2 = 18$ is choice (D).
Alternative: Instead of listing, factor $128 = 2^7$. The twin age appears squared, so it must take an even number of the seven $2$s: $2^0, 2^2, 2^4,$ or $2^6$, i.e. twin ages $1, 2, 4, 8$. Only $t = 8$ leaves Kiana ($2^1 = 2$) younger, again giving $8+8+2 = 18$.
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Turning the age story into the equation $t \times t \times k = 128$ with letters for the unknown ages.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing the whole-number twin ages whose square divides $128$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing each twin age with Kiana's age to keep only the case where the twins are older.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the three ages $8 + 8 + 2$ to get the final sum.)
⭐ When two people are the same age, that age gets used twice in the product, so only ages whose square fits the number can work.
⭐ When two people are the same age, that age gets used twice in the product, so only ages whose square fits the number can work.
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