AMC 10 · 2009 · #5
Easy mode Grade 4Kiana has two older brothers who are twins, so the two brothers are the same age. Both brothers are older than Kiana. When you multiply all three ages together, you get 128. What do the three ages add up to?
Pick an answer.
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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