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Difficulty: HardIntegers, Absolute Value, and Number Lines

Let SS be the set of all integers nn that satisfy the inequality ∣n−3∣+∣n+5∣≤12|n - 3| + |n + 5| \leq 12. What is the sum of the absolute values of all integers in set SS?

  1. 43Answer
  2. B
    -13
  3. C
    12
  4. D
    28
  5. E
    37

Answer

43
The correct answer is 43. Solving the inequality ∣n−3∣+∣n+5∣≤12|n - 3| + |n + 5| \leq 12 by casework on the number line yields the solution set of integers S={−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4,5}S = \{-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}. Summing the absolute values of these elements gives 7+6+5+4+3+2+1+0+1+2+3+4+5=437 + 6 + 5 + 4 + 3 + 2 + 1 + 0 + 1 + 2 + 3 + 4 + 5 = 43.

Step-by-Step Solution

1
Divide the number line into intervals to remove the absolute value bars based on critical points n=3n = 3 and n=−5n = -5.
Three cases to analyze: Case 1 (n≥3n \geq 3), Case 2 (−5≤n<3-5 \leq n < 3), and Case 3 (n<−5n < -5).
Evaluating absolute values requires knowing the signs of the expressions inside them.
2
Solve the inequality ∣n−3∣+∣n+5∣≤12|n - 3| + |n + 5| \leq 12 for each case.
For Case 1 (n≥3n \geq 3): (n−3)+(n+5)≤12⇒2n+2≤12⇒n≤5(n - 3) + (n + 5) \leq 12 \Rightarrow 2n + 2 \leq 12 \Rightarrow n \leq 5, yielding integers {3,4,5}\{3, 4, 5\}. For Case 2 (−5≤n<3-5 \leq n < 3): −(n−3)+(n+5)≤12⇒8≤12-(n - 3) + (n + 5) \leq 12 \Rightarrow 8 \leq 12 (always true), yielding integers {−5,−4,−3,−2,−1,0,1,2}\{-5, -4, -3, -2, -1, 0, 1, 2\}. For Case 3 (n<−5n < -5): −(n−3)−(n+5)≤12⇒−2n−2≤12⇒−2n≤14⇒n≥−7-(n - 3) - (n + 5) \leq 12 \Rightarrow -2n - 2 \leq 12 \Rightarrow -2n \leq 14 \Rightarrow n \geq -7, yielding integers {−7,−6}\{-7, -6\}.
Determining the integer values of nn that satisfy the inequality in each segment of the number line.
3
Combine the intervals to construct the complete set SS.
S={−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4,5}S = \{-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}.
The union of all valid cases gives the entire set of solutions.
4
Compute the sum of the absolute values of the integers in SS.
∣−7∣+∣−6∣+∣−5∣+∣−4∣+∣−3∣+∣−2∣+∣−1∣+∣0∣+∣1∣+∣2∣+∣3∣+∣4∣+∣5∣=7+6+5+4+3+2+1+0+1+2+3+4+5=43|-7| + |-6| + |-5| + |-4| + |-3| + |-2| + |-1| + |0| + |1| + |2| + |3| + |4| + |5| = 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 + 1 + 2 + 3 + 4 + 5 = 43.
Adding the absolute values of each element in the solution set.

Key Concept

Solving multi-interval absolute value inequalities with integers and finding absolute values
Estimated Time:2m 0s
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Topics

Integers, Absolute Value, and Number Lines

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