🇺🇸 AMC 8 ⇄ switch contest
1990 AJHSME

Problem 7

Problem 7 · 1990 AJHSME Medium
Arithmetic & Operations sign-productmaximize

When three different numbers from the set {−3, −2, −1, 4, 5} are multiplied, the largest possible product is

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Answer: C — 30.
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Hint 1 of 2
For the product to be as big as possible it must first be *positive*. With negatives around, what makes a product positive — how many minus signs do you need?
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Hint 2 of 2
An even number of negatives gives a positive product. So the play is: grab TWO negatives (to cancel the signs) and make them the biggest negatives you have, then pair with the biggest positive.
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Approach: even count of negatives for a positive, then maximize size
  1. First make sure you can even be positive. One negative would leave the product negative (a loser); using *two* negatives flips the signs to positive. So pick exactly two of the three negative numbers.
  2. To make the positive product as large as possible, use the two with the biggest size: (−3)(−2) = 6. Then multiply by the largest positive, 5.
  3. 6 × 5 = 30. (Check the rivals: 4×5 with one negative is negative; (−3)(−2)(4)=24 < 30.)
  4. *Why this transfers:* in any 'largest product' with negatives, the count of minus signs decides the sign first — settle the sign, *then* chase the biggest absolute values.
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