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Difficulty: Very hardConsecutive Integers and Number Sets

Set AA consists of kk consecutive integers, where k>1k > 1. Set BB consists of 2k2k consecutive integers. The median of Set AA is equal to the least integer in Set BB, and the sum of all integers in Set AA is equal to the sum of all integers in Set BB. If the sum of the integers in Set BB is −153-153, what is the greatest integer in Set AA?

  1. A
    −21-21
  2. B
    −17-17
  3. −13-13Answer
  4. D
    −12-12
  5. E
    −9-9

Answer

The greatest integer in Set AA is −13-13.
The problem establishes a relationship between two sets of consecutive integers using their medians and sums. Equating the sum formulas SA=k(a+k−12)S_A = k\left(a + \frac{k-1}{2}\right) and SB=2k(a+3k−22)S_B = 2k\left(a + \frac{3k-2}{2}\right) leads to a=3−5k2a = \frac{3-5k}{2}. Substituting this into the sum expression gives k−2k2=−153k - 2k^2 = -153, yielding k=9k = 9. Thus, the smallest element is a=−21a = -21, and the greatest element is −21+9−1=−13-21 + 9 - 1 = -13.

Step-by-Step Solution

1
Express the sum of Set AA in terms of kk and its smallest element aa.
The median of Set AA is MA=a+k−12M_A = a + \frac{k-1}{2}, so the sum of Set AA is SA=k⋅MA=k(a+k−12)S_A = k \cdot M_A = k\left(a + \frac{k-1}{2}\right).
In any set of consecutive integers, the sum equals the number of terms multiplied by the average (median).
2
Express the sum of Set BB using the given conditions.
The least integer in Set BB is MA=a+k−12M_A = a + \frac{k-1}{2}. Since Set BB has 2k2k consecutive integers, its median is MB=(a+k−12)+2k−12=a+3k−22M_B = \left(a + \frac{k-1}{2}\right) + \frac{2k-1}{2} = a + \frac{3k-2}{2}. The sum of Set BB is SB=2k⋅MB=2k(a+3k−22)=k(2a+3k−2)S_B = 2k \cdot M_B = 2k\left(a + \frac{3k-2}{2}\right) = k(2a + 3k - 2).
Set BB starts at the median of Set AA and contains 2k2k terms.
3
Equate SAS_A and SBS_B to solve for aa in terms of kk.
Since SA=SBS_A = S_B and k>1k > 1, we divide by kk: a+k−12=2a+3k−2  ⟹  a=3−5k2a + \frac{k-1}{2} = 2a + 3k - 2 \implies a = \frac{3 - 5k}{2}.
The problem states that the sums of the two sets are equal.
4
Substitute a=3−5k2a = \frac{3 - 5k}{2} into SBS_B and set it equal to −153-153.
SB=k(2(3−5k2)+3k−2)=k(3−5k+3k−2)=k(1−2k)=k−2k2S_B = k\left(2\left(\frac{3-5k}{2}\right) + 3k - 2\right) = k(3 - 5k + 3k - 2) = k(1 - 2k) = k - 2k^2. Setting k−2k2=−153  ⟹  2k2−k−153=0k - 2k^2 = -153 \implies 2k^2 - k - 153 = 0. Factoring gives (2k+17)(k−9)=0(2k + 17)(k - 9) = 0. Since kk must be a positive integer greater than 1, k=9k = 9.
Finding kk allows us to calculate the exact elements of Set AA.
5
Calculate the greatest integer in Set AA.
With k=9k = 9, the smallest integer a=3−5(9)2=−21a = \frac{3 - 5(9)}{2} = -21. The greatest integer in Set AA is a+k−1=−21+9−1=−13a + k - 1 = -21 + 9 - 1 = -13.
The kk-th term of a sequence of consecutive integers starting at aa is a+k−1a + k - 1.

Key Concept

Properties of consecutive integer sets: sum equals number of terms times median; indexing terms in evenly spaced sets.
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Topics

Consecutive Integers and Number Sets

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