Competition · AMC preparation · step 4 of 4
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.
Evaluate the inner box
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 the inner value
Substitute 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 SubproblemsEvaluate the outer box
Outer 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.
The positive factors of 12 are exactly 1, 2, 3, 4, 6, 12 — the list is complete, with nothing left out and nothing counted twice.
▸ Why?
A whole number counts as a factor of 12 exactly when it divides 12 with nothing left over, and dividing 12 evenly is the very same thing as saying some whole number multiplies back up to 12.
▸ Why?
No factor is missing, because the factors of 12 fall into partner pairs that each multiply to 12 — (1, 12), (2, 6), (3, 4) — so once the small partner of every pair is found, its large partner is pinned down as well.
▸ Why?
Matching each small factor with its partner 12 divided by it lines up the small factors 1, 2, 3 one for one with the large factors 12, 6, 4, so the small side and the large side are the same size and neither can hide an extra member.
▸ Why?
The largest factor cannot climb past 12 itself, since 12 equals 1 × 12 and 1 is the smallest partner a factor can have, so there is no factor above 12 to hunt for and only the candidates 1 through 12 need checking.
Read off the answer
Add 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.
- Evaluate the inner box
- Substitute the inner value
- Evaluate the outer box
- Read off the answer
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