AMC 8 · 2007 · #10
Grade 5 number-theoryPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Identify Subproblems) splits the nested expression 11 into two single-step questions: first 11 (which becomes a number), then that number. Each piece is just "list the factors and add them." Tool #2 (Make a Systematic List) is how we collect each factor exactly once — try divisors 1, 2, 3, … up to n in order, keep the ones that divide evenly. No algebra, no formulas; the two subproblems are both Grade 4 work.
Inner box first: 11 is prime, so its only factors are 1 and 11, which add to 12.
A prime number has exactly two factors: 1 and itself. That makes the inner box the easiest possible case.
4.OA.B.4Make A Systematic ListSubstitute inside-out: the inner box is 12, so the problem is now the box of 12.
Working from the inside of a nested expression outward is the standard "parentheses first" order-of-operations move.
5.OA.A.1Identify SubproblemsOuter box: list the divisors of 12 in order — 1, 2, 3, 4, 6, 12 (5 is skipped since it does not divide 12).
Listing factors in order from 1 upward guarantees none are missed. Stopping at √(12) ≈ 3.5 and pairing each small factor with 12 ÷ factor also gives the same six numbers.
4.OA.B.4Make A Systematic ListAdd those six factors: the outer box equals 28 — choice (D).
Adding the six factors of 12 in order: 1+2 = 3, 3+3 = 6, 6+4 = 10, 10+6 = 16, 16+12 = 28. Choice (D).
4.NBT.B.4Identify SubproblemsWhen a symbol is wrapped inside itself, evaluate the inside first and substitute — then this AMC 8 problem is just two Grade 4 factor lists in a row.
More like this
Same archetype — closest grade level first.
