SOLUTION: there are three consecutive even integers. if twice the first integer added to the second is 268,214, find all three integers.

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Question 1196211: there are three consecutive even integers. if twice the first integer added to the second is 268,214, find all three integers.
Found 3 solutions by ikleyn, MathLover1, greenestamps:
Answer by ikleyn(52825)   (Show Source): You can put this solution on YOUR website!
.

The numbers are n, (n+2) and (n+4), from smaller to greater.


The equation is

    2n + (n+2) = 268214.


Solve it step by step

    3n = 268214 - 2 = 268212.

     n              = 268212/3 = 89404.


ANSWER.  The numbers are 89404, 89406 and 89408.

Solved.

---------------

After my post,  @MathLover1  came with her own solution,
but the numbers in her answer  ( 8942,  9422,  89424 )  are incorrect.



Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

I will answer it as stated.
three consecutive even integers are , ,

if twice the first integer added to the second is , we have






Then the three consecutive even integers are ,,


Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


Note that since the numbers are "very close together", you can solve the problem informally with simple arithmetic and a bit of logical reasoning.

Twice the first integer added to the second integer is very nearly the same as three times either integer. So divide the total 268214 by 3 to get 89404 2/3. Then, since you know the actual numbers are three consecutive even integers, the first and second numbers must be 89404 and 89406.

Check to make sure they work: 2(89404)+89406=178808+89406=268214.

ANSWER: the three consecutive even integers are 89404, 89406, and 89408.
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