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Difficulty: Very hardIntegers, Absolute Value, and Number Lines

Let xx, yy, and zz be integers such that ∣x∣<∣y∣<∣z∣|x| < |y| < |z|, x+y+z=−3x + y + z = -3, and xyz>0xyz > 0. What is the smallest possible value of ∣x−y∣+∣y−z∣+∣z−x∣|x - y| + |y - z| + |z - x|?

  1. A
    6
  2. B
    8
  3. C
    10
  4. 12Answer
  5. E
    14

Answer

12
The correct answer is 12. To satisfy the given conditions, the product xyz>0xyz > 0 means either all three numbers are positive, or exactly two are negative. Since they sum to −3-3, they cannot all be positive. Letting two be negative and one positive, we find that the integer set {−1,2,−4}\{-1, 2, -4\} satisfies ∣−1∣<∣2∣<∣−4∣|-1| < |2| < |-4| (or 1<2<41 < 2 < 4), has a sum of −3-3, and a product of 8>08 > 0. The sum of the absolute differences is ∣−1−2∣+∣2−(−4)∣+∣−4−(−1)∣=3+6+3=12|-1 - 2| + |2 - (-4)| + |-4 - (-1)| = 3 + 6 + 3 = 12, which is the minimum possible value.

Step-by-Step Solution

1
Analyze the sign constraints from the product condition xyz>0xyz > 0.
Since the product of xx, yy, and zz is positive, either all three integers are positive, or exactly two are negative and one is positive.
The product of three numbers is positive if there are zero or two negative factors.
2
Evaluate the case where all three integers are positive.
If all three are positive, then x,y,z≥1x, y, z \geq 1. This implies their sum x+y+z≥3x+y+z \geq 3, which contradicts the constraint x+y+z=−3x+y+z = -3. Thus, this case is impossible.
Positive integers cannot sum to a negative number.
3
Analyze the case where two integers are negative and one is positive, and express them in terms of their absolute values.
Let the negative integers be −u-u and −v-v (with u>v>0u > v > 0) and the positive integer be p>0p > 0. The sum constraint becomes −u−v+p=−3-u - v + p = -3, which simplifies to p=u+v−3p = u + v - 3. The absolute values are uu, vv, and pp, which must be three distinct positive integers satisfying ∣x∣<∣y∣<∣z∣|x| < |y| < |z|.
This translates the constraints into positive integer variables representing the absolute values.
4
Determine the values of u,v,pu, v, p that minimize the maximum difference between any two of the three integers x,y,zx, y, z.
For v=1v = 1: if u=4u = 4, then p=2p = 2. The absolute values {1,2,4}\{1, 2, 4\} are distinct. The corresponding integers are x=−1x = -1, y=2y = 2, and z=−4z = -4. Their sum is −3-3 and product is 8>08 > 0. The difference between the maximum and minimum values is 2−(−4)=62 - (-4) = 6. For v=2v = 2: if u=4u = 4, then p=3p = 3. The absolute values {2,3,4}\{2, 3, 4\} are distinct, giving the integers −4,−2,3-4, -2, 3, with a maximum difference of 3−(−4)=73 - (-4) = 7. For higher values, the difference only increases.
The expression ∣x−y∣+∣y−z∣+∣z−x∣|x - y| + |y - z| + |z - x| is equal to twice the difference between the maximum and minimum of the three integers.
5
Calculate the minimum value of the expression.
Using the optimal set x=−1,y=2,z=−4x = -1, y = 2, z = -4, the value of the expression is ∣−1−2∣+∣2−(−4)∣+∣−4−(−1)∣=3+6+3=12|-1 - 2| + |2 - (-4)| + |-4 - (-1)| = 3 + 6 + 3 = 12.
This is the smallest possible sum of the absolute differences.

Key Concept

Using absolute values and sign analysis to solve system of constraints on integers and distances on a number line.
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Topics

Integers, Absolute Value, and Number Lines

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