AMC 8 · 2007 · #10
Easy mode Grade 5Problem
We are going to use a special symbol. For any positive whole number n, the symbol n means "add up every positive factor of n."
For example, the factors of 6 are 1,2,3, and 6. So 6=1+2+3+6=12.
Now we want to find 11.
To do this, first compute 11. Then apply the box symbol again to that result.
Pick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Define $\boxed{n}$ to be the sum of all positive factors of $n$. For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12$. Compute $\boxed{\boxed{11}}$.
Givens: $\boxed{n}$ means the sum of the positive factors of $n$; Example: $\boxed{6} = 1 + 2 + 3 + 6 = 12$; We must evaluate $\boxed{\boxed{11}}$ — the same operation applied twice, starting at $11$; Answer choices: (A) $13$, (B) $20$, (C) $24$, (D) $28$, (E) $30$
Unknowns: The value of $\boxed{\boxed{11}}$
Understand
Restated: Define $\boxed{n}$ to be the sum of all positive factors of $n$. For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12$. Compute $\boxed{\boxed{11}}$.
Givens: $\boxed{n}$ means the sum of the positive factors of $n$; Example: $\boxed{6} = 1 + 2 + 3 + 6 = 12$; We must evaluate $\boxed{\boxed{11}}$ — the same operation applied twice, starting at $11$; Answer choices: (A) $13$, (B) $20$, (C) $24$, (D) $28$, (E) $30$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
Tool #7 (Identify Subproblems) splits the nested expression $\boxed{\boxed{11}}$ into two single-step questions: first $\boxed{11}$ (which becomes a number), then $\boxed{\text{that number}}$. 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 $1, 2, 3, \ldots$ up to $n$ in order, keep the ones that divide evenly. No algebra, no formulas; the two subproblems are both Grade 4 work.
Execute — Answer: D
4.OA.B.4 Step 1 - Evaluate the inner box: $\boxed{11}$.
- List every positive factor of $11$.
- Since $11$ is prime, only $1$ and $11$ divide it evenly.
💡 A prime number has exactly two factors: $1$ and itself. That makes the inner box the easiest possible case.
5.OA.A.1 Step 2 - Replace the inner box with its value.
- The expression $\boxed{\boxed{11}}$ now becomes $\boxed{12}$.
💡 Working from the inside of a nested expression outward is the standard "parentheses first" order-of-operations move.
4.OA.B.4 Step 3 - Evaluate the outer box: $\boxed{12}$.
- Walk the divisors $1, 2, 3, 4, 5, 6, \ldots$ in order and keep the ones that divide $12$ evenly.
- The factors of $12$ are $1, 2, 3, 4, 6, 12$ (note $5$ does not divide $12$; after $6$ the next factor would be $12 \div 1 = 12$).
💡 Listing factors in order from $1$ upward guarantees none are missed. Stopping at $\sqrt{12} \approx 3.5$ and pairing each small factor with $12 \div \text{factor}$ also gives the same six numbers.
4.NBT.B.4 Step 4 The outer sum is the answer.
💡 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.OA.B.4 Evaluate the inner box: $\boxed{11}$. List every positive factor of $11$. Since 5.OA.A.1 Replace the inner box with its value. The expression $\boxed{\boxed{11}}$ now be 4.OA.B.4 Evaluate the outer box: $\boxed{12}$. Walk the divisors $1, 2, 3, 4, 5, 6, \ldot 4.NBT.B.4 The outer sum is the answer. Review
Reasonableness: Re-pair the factors of $12$ to double-check the sum: $1 \cdot 12 = 12$, $2 \cdot 6 = 12$, $3 \cdot 4 = 12$. Three pairs that each multiply to $12$, with sums $1+12 = 13$, $2+6 = 8$, $3+4 = 7$. Total: $13 + 8 + 7 = 28$. The answer (D) matches, and the size feels right — $\boxed{12}$ should be a bit more than $12$ itself (since the proper factors $1, 2, 3, 4, 6$ alone already add to $16$). Also, $5$ correctly drops out because $12 / 5$ is not a whole number.
Alternative: Tool #3 (Eliminate Possibilities): once $\boxed{11} = 12$ is in hand, the question reduces to "sum of factors of $12$." The factors must include $1$ and $12$ themselves, so the total is at least $13$. That already rules out (A) $13$ (which would force $12$ to be prime — false, since $2$ divides $12$). Adding the obvious divisors $2$ and $3$ pushes the total to at least $1 + 2 + 3 + 12 = 18$. Adding $4$ and $6$ brings it to $28$. (D) is the only choice consistent with all six factors.
CCSS standards used (min grade 5)
4.OA.B.4Find factor pairs, recognize multiples, and identify prime numbers within $1$-$100$ (Listing the factors of $11$ (just $1$ and $11$, since $11$ is prime) and the factors of $12$ ($1, 2, 3, 4, 6, 12$) — the Grade 4 factor-pairs and primes standard.)5.OA.A.1Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols (Evaluating the nested expression $\boxed{\boxed{11}}$ from the inside out — the Grade 5 grouping-symbols standard governs this innermost-first order.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Summing the six factors of $12$: $1 + 2 + 3 + 4 + 6 + 12 = 28$.)
⭐ When 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.
⭐ When 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.
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