Problem 21 · 2017 AMC 8
Hard
Algebra & Patterns
caseworksubstitution
Suppose a, b, and c are nonzero real numbers, and a + b + c = 0. What are the possible value(s) for
a|a| + b|b| + c|c| + abc|abc| ?
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Answer: A — 0.
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Hint 1 of 2
Each x/|x| is just a sign: +1 if positive, −1 if negative. So the whole expression is (sign of a) + (sign of b) + (sign of c) + (sign of abc).
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Hint 2 of 2
The constraint a+b+c=0 forbids all-same-sign, so the count of negatives is exactly 1 or 2. Notice the last term's sign is determined by the first three: an odd number of negatives makes abc negative.
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Approach: read everything as signs (+1/−1)
- Each x/|x| equals +1 (if x>0) or −1 (if x<0). So we're adding four signs — three for a, b, c, and one for their product abc.
- Since the three sum to 0 and none is zero, they aren't all the same sign: exactly one or exactly two of a, b, c are negative.
- Two positive, one negative: the three signs are +1, +1, −1 (sum +1); one negative factor makes abc negative, sign −1. Total: +1 − 1 = 0.
- Two negative, one positive: signs −1, −1, +1 (sum −1); two negatives make abc positive, sign +1. Total: −1 + 1 = 0.
- Both cases give 0. The deeper reason: the abc sign always cancels the sum of the first three, because the product's sign tracks the parity of how many are negative.
Another way — pick a concrete example to kill wrong choices:
- The value can't depend on which specific numbers you pick, so test one: a=1, b=1, c=−2 (sum 0). Then signs give 1 + 1 + (−1) + (−1) = 0.
- A single value of 0 already eliminates every choice except (A); the casework above confirms 0 is the only value, so the answer is 0.
- Test-taking tip: when a problem asks 'what are the possible values' and the answer must be constant, computing one clean example often pins it instantly.
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