SOLUTION: Find three consecutive integers such that three times the smallest integer, increased by the largest integer equals 106.

Question 648133: Find three consecutive integers such that three times the smallest integer, increased by the largest integer equals 106.
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Consecutive interger or odd interger or even integer problems are straight forward once you get the hang of representing them as an algebraic expression. I always let the first (smallest) integer be defined as
(1) n = the first (smallest) integer.
Then the next consecutive integer is
(2) (n + 1) = next (or second) consecutive integer
Note here, that if you want the next conseutive odd or even integer it would be (n + 2); that is we skip the next integer (like counting by two).
And the third consecutive integer is (n+1)+1 or
(3) (n + 2) = the third (and largest in this problem) consecutive integer
Make sense?
Now write an expression using these integer definitions that satisfy the problem statement.
Convert,"three times the smallest integer (n) increased (add) by the largest (n+2) is equal to 106," into the following equality
(4) 3*n + (n + 2) = 106
Simplify (4) to yield
(5) 4*n = 104
Divide each side of (5) by 4 to yield
(6) n = 26
Then the sequence of three consecutive integers is 26,27,28.
Always check your answer. Let's do this by checking (4).
Is (3*26 + 28 = 106)?
Is (78 + 28 = 106)?
Is (106 = 106)? Yes
Answer: The three consecutive integers are 26,27 and 28.
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