Question

Given three consecutive odd integers, the sum of the smallest integer and twice the largest integer is ten squared more than the median integer. What is the largest integer?

a. 57

b. 61

c. 65

d. 69

Answer 1

To solve this problem, assume the three consecutive odd integers are represented by n, n+2, and n+4. Write equations based on the given information and solve to find n and the largest integer.

Step-by-step explanation:

To solve this problem, let's assume that the three consecutive odd integers are represented by n, n+2, and n+4. The sum of the smallest integer and twice the largest integer is n + 2(n+4). The median integer is n + 2. The problem states that this sum is ten squared more than the median integer. So we can write the equation as: n + 2(n+4) = (n + 2)^2 + 10^2.

Expanding and simplifying the equation gives us: n + 2n + 8 = n^2 + 4n + 4 + 100.

Combining like terms gives us: 3n + 8 = n^2 + 4n + 104.

Bringing all terms to one side of the equation gives us: n^2 + n - 96 = 0.

Factoring the quadratic equation gives us: (n - 8)(n + 12) = 0.

So n = 12 or n = -8. Since we are dealing with odd integers, n must be 12. Therefore, the largest integer is n + 4 = 12 + 4 = 16. So the correct option is d. 69.