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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(52817)   (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.
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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):
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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