AMC 10 · 2024 · #7

Grade 7 number-theoryarithmetic

Archetype Small-Domain Constrained Search

factorsparitysystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticprime-factorization

📏 Medium solution 💡 3 insights

Problem

The product of three integers is $60$ . What is the least possible positive sum of the
three integers?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }13$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Three integers multiply to $60$. Among all such triples, find the smallest sum that is still positive.

Givens: Three integers $a, b, c$ with $a \cdot b \cdot c = 60$; Integers may be positive or negative (the word "integer" is not restricted to positive); Answer choices: (A) $2$, (B) $3$, (C) $5$, (D) $6$, (E) $13$

Unknowns: The least positive value of $a + b + c$

Understand

Restated: Three integers multiply to $60$. Among all such triples, find the smallest sum that is still positive.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List

The sign rule for a positive product splits the search cleanly into two subproblems (Tool #7): Case A — all three factors positive; Case B — one positive and two negative. Each case has only finitely many factor triples of $60$, so Tool #2 (Make a Systematic List) finishes the work: list the triples in order, compute each sum, keep the smallest positive value. Comparing the two cases gives the overall minimum.

Execute — Answer: B

#7 Identify Subproblems 7.NS.A.2 Step 1

Split by sign.
A product of three integers equals $60 > 0$ only when an even number of the factors are negative.
With three integers that means $0$ negatives (all positive) or $2$ negatives (one positive, two negative).
Handle each case separately.

$$\text{Case A: } a,b,c > 0. \qquad \text{Case B: } a > 0,\; b,c < 0.$$

💡 Grade 7 sign rules for multiplication: negative $\times$ negative $=$ positive, so two negatives cancel and the product stays positive.

#2 Make a Systematic List 6.NS.B.4 Step 2

Case A: list factor triples of $60$ in non-decreasing order.
To keep the sum small, push the factors close together.
Walk through the small first factor systematically.

$$1 \cdot 1 \cdot 60 = 60,\; \text{sum } 62.\;\; 1 \cdot 2 \cdot 30 = 60,\; \text{sum } 33.\;\; 1 \cdot 3 \cdot 20,\; 24.\;\; 1 \cdot 4 \cdot 15,\; 20.\;\; 1 \cdot 5 \cdot 12,\; 18.\;\; 1 \cdot 6 \cdot 10,\; 17.\;\; 2 \cdot 2 \cdot 15,\; 19.\;\; 2 \cdot 3 \cdot 10,\; 15.\;\; 2 \cdot 5 \cdot 6,\; 13.\;\; 3 \cdot 4 \cdot 5,\; 12.$$

💡 Grade 6 factor work: walk the first factor up from $1$, and for each first factor list the matching pairs of the leftover quotient.

#2 Make a Systematic List 6.NS.B.4 Step 3

Read off the Case A minimum.
The smallest sum in the list is $3 + 4 + 5 = 12$, reached when the factors are as close together as possible.

$$\min_{\text{Case A}}(a+b+c) = 3+4+5 = 12.$$

💡 For a fixed product, sums shrink as factors approach each other (AM-GM intuition you can see by scanning the list).

#2 Make a Systematic List 7.NS.A.1 Step 4

Case B: write the triple as $(p, -q, -r)$ with $p,q,r > 0$ and $pqr = 60$.
The sum becomes $p - q - r$.
To make this a small positive number, pick $p$ slightly larger than $q + r$.
Reuse the same factor-triple list, this time assigning the largest factor to $p$.

$$\text{Sum } = p - (q+r),\;\; \text{need } p > q+r \text{ for positivity.}$$

💡 Grade 7 integer addition: combining a positive with two negatives is just $p$ minus the total of the two absolute values.

#2 Make a Systematic List 7.NS.A.1 Step 5

Scan Case B candidates.
For each triple, set $p$ = largest, check $p - q - r$.
Track only positive sums and keep shrinking.
- $\{1,1,60\}$: $60-1-1 = 58$.
- $\{1,2,30\}$: $30-1-2 = 27$.
- $\{1,3,20\}$: $20-1-3 = 16$.
- $\{1,4,15\}$: $15-1-4 = 10$.
- $\{1,5,12\}$: $12-1-5 = 6$.
- $\{1,6,10\}$: $10-1-6 = 3$.
(positive and small) - $\{2,2,15\}$: $15-2-2 = 11$.
- $\{2,3,10\}$: $10-2-3 = 5$.
- $\{2,5,6\}$: $6-2-5 = -1$ (not positive, skip).
- $\{3,4,5\}$: $5-3-4 = -2$ (not positive, skip).

$$\min_{\text{Case B, positive}}(p-q-r) = 10 - 6 - 1 = 3,\;\; \text{from } (10, -6, -1).$$

💡 Once $p$ falls below $q+r$ the sum turns negative, so the smallest positive answer hides right at the boundary $p \approx q+r$.

#7 Identify Subproblems 6.NS.C.7 Step 6

Compare the two cases.
Case A bottoms out at $12$, Case B bottoms out at $3$.
The overall least positive sum is $3$, matching choice (B).

$$\min(12,\;3) = 3 \;\Rightarrow\; \textbf{(B)}.$$

💡 Grade 6 ordering of integers: just compare two values and pick the smaller one.

[1] #7 7.NS.A.2 Split by sign. A product of three integers equals $60 > 0$ only when an even num

[2] #2 6.NS.B.4 Case A: list factor triples of $60$ in non-decreasing order. To keep the sum sma

[3] #2 6.NS.B.4 Read off the Case A minimum. The smallest sum in the list is $3 + 4 + 5 = 12$, r

[4] #2 7.NS.A.1 Case B: write the triple as $(p, -q, -r)$ with $p,q,r > 0$ and $pqr = 60$. The s

[5] #2 7.NS.A.1 Scan Case B candidates. For each triple, set $p$ = largest, check $p - q - r$. T

[6] #7 6.NS.C.7 Compare the two cases. Case A bottoms out at $12$, Case B bottoms out at $3$. Th

Review

Reasonableness: Verify the winning triple $(10, -6, -1)$ end to end. Product: $10 \cdot (-6) \cdot (-1) = 10 \cdot 6 = 60$, matches. Sum: $10 + (-6) + (-1) = 10 - 7 = 3$, positive as required. Could a sum of $1$ or $2$ exist? Any Case-B triple needs $p - q - r \in \{1, 2\}$, i.e. $p = q + r + 1$ or $p = q + r + 2$, with $pqr = 60$. Checking divisor triples of $60$ exhausts the options: $(60,1,1)$ gives $58$, $(30,2,1)$ gives $27$, $(20,3,1)$ gives $16$, $(15,4,1)$ gives $10$, $(12,5,1)$ gives $6$, $(10,6,1)$ gives $3$, then everything else falls below $3$ or below $0$. So $3$ is truly the minimum and (B) is locked in.

Alternative: Tool #5 (Look for a Pattern): order the divisor triples of $60$ by their largest factor $p$. The Case-B sum $p-q-r$ drops monotonically as $p$ shrinks — until it crosses zero. The last positive value before the crossover is the minimum. That "last positive before zero" is $(10,6,1) \to 3$, so the answer is (B) without exhaustive search.

CCSS standards used (min grade 7)

7.NS.A.2 Apply and extend previous understandings of multiplication and division to multiply and divide rational numbers (Splitting into sign cases: a product of three integers is positive only when zero or two of the factors are negative.)
6.NS.B.4 Find the greatest common factor and least common multiple; list and use factors (Systematically listing the unordered factor triples of $60$ in both cases.)
7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Computing $p + (-q) + (-r) = p - q - r$ for the mixed-sign case and tracking when that value stays positive.)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Comparing the two case minima ($12$ vs. $3$) to pick the smaller.)

⭐ Negative times negative is positive, so two negative factors can cancel — that lets the giant factor $10$ shrink down to a sum of just $3$. Splitting into sign cases first, then listing factor triples second, turns this AMC 10 question into ordinary Grade 6-7 work.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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