SOLUTION: The sum of three consecutive integers is at most 442 less twice the largest integer. What are the integers?

Question 1170108: The sum of three consecutive integers is at most 442 less twice the largest integer. What are the integers?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let the smallest integer be equal to x.
let the next larger integer be equal to x + 1.
let the largest integer be equal to x + 2.

you are given that the sum of the 3 consecutive integers is at most equal to 442 less than twice the largest integer.

that means that x + x + 1 + x + 2 <= 2 * (x + 2) - 442.

combine like terms on the left side of the inequality and simplify the right side of the inequality to get:

3x + 3 <= 2x - 438.

subtract 3 from both sides of the inequality and subtract 2x from both sides of the inequality and simplify to get:

x <= -441.

when x = -441, then the three consecutive integers are -441, -440, -439.

their sum is equal to -1320.

two times the largest integer is equal to -878.

the sum of the three consecutive integers is equal to -1320.

-1320 - -878 = (-1320 + 878) = 442.

this means -1320 is 442 less than -878.

this satisfies the statement that the sum of the three consecutive integers is at most 442 less than twice the largest integer.

your solution appears to be that the integers are -441, -440, -439.

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