AMC 10 · 2009 · #6
Grade 4 number-theoryPick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Name the twin age t and Kiana's age k. Since the twins share an age, the product becomes t × t × 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.
Name the ages
Let t be the twins' shared age and k be Kiana's age. The twins match, so t × t × k = 128.
Using one letter for the twins captures the fact that their two ages are really the same number.
4.OA.A.3Introduce A VariableList the twin ages that fit
Since 128 is 2 multiplied seven times, t can only be 1, 2, 4, or 8, and then k = 128 ÷ t² is 128, 32, 8, or 2.
Since the twin age is used twice, only ages whose square fits inside 128 are even possible.
Since the twin age is used twice, only ages whose square fits inside the product are even possible.
▸ Why?
The product has exactly one prime recipe, so which factors can appear is settled in advance.
▸ Why?
Divisors come in pairs that multiply back to the number, so the candidate ages form a short closed list.
Keep only older twins
The brothers are older, so t must beat k. Of the four cases only t = 8 with 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.A.2Eliminate PossibilitiesAdd the three ages
The three ages are 8, 8, and 2, so the sum is 2t + k = 18, which is choice (D).
Once the only valid ages are found, adding them is the last easy step.
4.NBT.B.4Introduce A VariableWhen two people are the same age, that age gets used twice in the product, so only ages whose square fits the number can work.
- Name the ages
- List the twin ages that fit
- Keep only older twins
- Add the three ages