AMC 10 · 2008 · #3
Grade 4 number-theoryPick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The triple brackets are three separate ⟨ ⟩ operations stacked on each other. Evaluate the innermost one first, then feed its result into the next. Each ⟨ ⟩ is itself a small task: list the proper divisors, then add them.
List the divisors of 6
Inside the brackets, 6 splits as 1 × 6 and 2 × 3, so its divisors are 1, 2, 3, 6 — drop 6 itself and keep 1, 2, 3.
Every divisor pairs with a partner that multiplies back to 6, so checking up to 3 already finds them all.
Every divisor pairs with a partner that multiplies back to the number, so checking halfway finds them all.
▸ Why?
Divisors come in such pairs, so listing the small side lists the large side too.
▸ Why?
Every number has exactly one prime recipe, which fixes the whole list of its divisors in advance.
Add them to get the value of 6
Adding them closes the innermost bracket: 1 + 2 + 3 = 6, so ⟨ 6⟩ = 6 — the number returns itself, a perfect number.
A perfect number equals the sum of its own proper divisors, so ⟨ 6⟩ lands right back on 6.
1.OA.C.6Identify SubproblemsApply the operation again
Since ⟨ 6⟩ is 6, the middle bracket just reads ⟨ 6⟩ all over again, so ⟨⟨ 6⟩⟩ = 6 — the value refuses to move.
Once a number maps to itself, repeating the rule keeps handing back the same number.
4.OA.C.5Look For A PatternApply the operation a third time
The outer bracket repeats it once more: ⟨⟨⟨ 6⟩⟩⟩ = ⟨ 6⟩ = 6, so any depth of nesting still gives answer (A).
6 is a fixed point of ⟨ ⟩, so nesting the brackets to any depth still gives 6.
4.OA.C.5Look For A Pattern6's proper divisors add back up to 6, so no matter how many times you take the divisor-sum, you always land on 6.
- List the divisors of 6
- Add them to get the value of 6
- Apply the operation again
- Apply the operation a third time