AMC 8 · 2000 · #7
Grade 7 arithmeticProblem
What is the minimum possible product of three different numbers of the set {−8,−6,−4,0,3,5,7}?
Pick an answer.
AMC 8 2000 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: From the set $\{-8, -6, -4, 0, 3, 5, 7\}$, pick three different numbers and multiply them. What is the smallest (most negative) product you can make?
Givens: The set is $\{-8, -6, -4, 0, 3, 5, 7\}$ — three negatives, one zero, three positives; Exactly three different numbers are multiplied; Answer choices: (A) $-336$, (B) $-280$, (C) $-210$, (D) $-192$, (E) $0$
Unknowns: The minimum possible product of three distinct chosen numbers
Understand
Restated: From the set $\{-8, -6, -4, 0, 3, 5, 7\}$, pick three different numbers and multiply them. What is the smallest (most negative) product you can make?
Givens: The set is $\{-8, -6, -4, 0, 3, 5, 7\}$ — three negatives, one zero, three positives; Exactly three different numbers are multiplied; Answer choices: (A) $-336$, (B) $-280$, (C) $-210$, (D) $-192$, (E) $0$
Plan
Primary tool: #14 Extreme Principle
Secondary: #2 List Out Cases
We want the most negative product, so Tool #14 (Extreme Principle) applies: the answer comes from pushing the chosen numbers to extreme magnitudes with the right signs. The sign rules narrow the search: three numbers multiply to a negative only when the count of negatives is $1$ or $3$, so Tool #2 (List Out Cases) gives just two cases to compare. Inside each case, the Extreme Principle picks the numbers with the largest absolute values. Skip $0$ — any product containing it is $0$, which is not the smallest.
Execute — Answer: B
7.NS.A.2 Step 1 - List the sign patterns of three numbers that produce a negative product.
- With three factors, the product is negative exactly when an odd number of them are negative — so either $1$ negative or $3$ negatives.
- Including $0$ makes the product $0$, which is larger than any negative, so $0$ is out.
💡 The Grade 7 sign rule for multiplication says each negative factor flips the sign once. An odd number of flips leaves the product negative.
7.NS.A.2 Step 2 - Case A — pick all three negatives.
- There is only one way: use $-8$, $-6$, $-4$.
- Two negatives multiply to a positive, then a third negative flips back to negative.
💡 No choice to make in this case — the three negatives are fixed, so the product is forced to $-192$.
6.NS.C.7 Step 3 - Case B — pick $1$ negative and $2$ positives.
- To make the product as negative as possible, push the absolute value to its maximum.
- The Extreme Principle says: choose the negative with the largest absolute value, and the two largest positives.
💡 Bigger magnitudes on each factor make the size of the product bigger. Since the sign is locked negative by the one negative factor, bigger size means more negative.
6.NS.C.7 Step 4 - Compute Case B and compare with Case A.
- The most-negative product in Case B is $(-8) \times 7 \times 5 = -280$.
- Compare with Case A's $-192$.
- On the number line, $-280 < -192$, so Case B wins.
💡 Comparing two negative numbers: the one farther from $0$ is smaller. $-280$ is farther left on the number line than $-192$.
7.NS.A.2 List the sign patterns of three numbers that produce a negative product. With th 7.NS.A.2 Case A — pick all three negatives. There is only one way: use $-8$, $-6$, $-4$. 6.NS.C.7 Case B — pick $1$ negative and $2$ positives. To make the product as negative as 6.NS.C.7 Compute Case B and compare with Case A. The most-negative product in Case B is $ Review
Reasonableness: Spot-check by trying other Case B picks. Using $-6$ instead of $-8$: $(-6) \times 7 \times 5 = -210$ (choice C) — less negative than $-280$, so $-8$ was the right pick. Using $-8$ but with $7$ and $3$: $(-8) \times 7 \times 3 = -168$ — less negative. Choice (A) $-336$ would need $|-8| \times |\text{two numbers}| = 336$, i.e. two positives whose product is $42$ — but the largest available pair gives $7 \times 5 = 35$, not $42$, so $-336$ is unreachable. The answer (B) $-280$ holds.
Alternative: Tool #3 (Eliminate Possibilities): choice (E) $0$ is not the smallest because we can reach negatives. Choice (A) $-336 = -8 \times 42$, but no two distinct positives in the set multiply to $42$ ($7 \times 5 = 35$ is the maximum). That removes (A) and (E). Among (B), (C), (D), the Case A all-negatives product is exactly $-192$ (D), and Case B's best is $-280$ (B), which beats $-210$ (C). So (B) wins.
CCSS standards used (min grade 7)
7.NS.A.2Apply and extend previous understandings of multiplication and division to multiply and divide rational numbers, including signed numbers (Using the sign rule for products to see that a negative product needs an odd count of negative factors, and computing $(-8) \times (-6) \times (-4) = -192$ and $(-8) \times 7 \times 5 = -280$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Comparing $-280$ and $-192$ on the number line (the one with larger absolute value is the smaller number) and choosing the negative with the largest absolute value to maximize the size of the product.)
⭐ For a smallest-product question, list only the sign patterns that go negative, then within each one make every factor as big in size as possible. Two short cases beat any guess-and-shuffle approach.
⭐ For a smallest-product question, list only the sign patterns that go negative, then within each one make every factor as big in size as possible. Two short cases beat any guess-and-shuffle approach.
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