AMC 8 · 2000 · #7
Grade 7 arithmeticPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
The product is negative only when an odd number of factors are negative — 1 or 3 negatives; any pick with 0 gives 0, so drop 0.
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.2Make A Systematic ListCase A takes all three negatives -8, -6, -4: two negatives make a positive, the third flips it back to -192.
No choice to make in this case — the three negatives are fixed, so the product is forced to -192.
7.NS.A.2Make A Systematic ListCase B (one negative, two positives): the Extreme Principle takes the biggest-size negative and two biggest positives — -8, 7, 5.
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.7Evaluate Finite DifferencesCase B gives (-8)×7×5 = -280; since -280 < -192 on the number line, the minimum is -280, choice (B).
Comparing two negative numbers: the one farther from 0 is smaller. -280 is farther left on the number line than -192.
6.NS.C.7Evaluate Finite DifferencesFor 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.