Question

Difficulty: MediumConsecutive Integers and Number Sets

Set SS consists of consecutive integers from aa to bb, inclusive, such that a<0<ba < 0 < b. The number of negative integers in set SS is equal to three times the number of positive integers in set SS. If the sum of all elements in set SS is 68-68, how many integers are in set SS?

  1. A
    15
  2. B
    16
  3. 17Answer
  4. D
    21
  5. E
    13

Answer

17
Set SS consists of 1212 negative integers (12-12 through 1-1), 44 positive integers (11 through 44), and the integer 00. The sum of positive integers is 1010 and the sum of negative integers is 78-78, giving a total sum of 68-68. The total number of integers in the set is 12+4+1=1712 + 4 + 1 = 17.

Step-by-Step Solution

1
Define the number of terms and bounds of set SS in terms of positive integers bb.
The positive integers are 1,2,,b1, 2, \dots, b, so there are bb positive integers. The negative integers are 1,2,,a-1, -2, \dots, a. Since the number of negative integers is 33 times the number of positive integers, there are 3b3b negative integers, which means a=3ba = -3b. Set SS also includes 00.
Expressing all elements in terms of bb allows setting up a single variable equation.
2
Calculate the sum of all elements in set SS as a function of bb.
The sum of the positive integers is b(b+1)2\frac{b(b+1)}{2}. The sum of the negative integers is 3b(3b+1)2-\frac{3b(3b+1)}{2}. The total sum is b(b+1)23b(3b+1)2=b2+b9b23b2=8b22b2=4b2b\frac{b(b+1)}{2} - \frac{3b(3b+1)}{2} = \frac{b^2 + b - 9b^2 - 3b}{2} = \frac{-8b^2 - 2b}{2} = -4b^2 - b.
The sum of consecutive integers from 11 to nn is n(n+1)2\frac{n(n+1)}{2}.
3
Solve for bb using the given sum of 68-68.
4b2b=68    4b2+b68=0-4b^2 - b = -68 \implies 4b^2 + b - 68 = 0. Factoring gives (4b+17)(b4)=0(4b + 17)(b - 4) = 0. Since bb must be a positive integer, b=4b = 4. Therefore, a=3(4)=12a = -3(4) = -12.
Solving the quadratic equation identifies the exact boundaries of the set.
4
Calculate the total number of integers in set SS.
The set contains integers from 12-12 to 44, inclusive. The total count of integers is 4(12)+1=174 - (-12) + 1 = 17.
The number of integers in an inclusive range [a,b][a, b] is given by ba+1b - a + 1.

Key Concept

Consecutive Integers and Number Sets
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