SOLUTION: One half the sum of three consecutive numbers added to twice the first number is equal to 63, find the smallest integer

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Question 989185: One half the sum of three consecutive numbers added to twice the first number is equal to 63, find the smallest integer
Found 4 solutions by CubeyThePenguin, MathTherapy, josgarithmetic, ikleyn:
Answer by CubeyThePenguin(3113) About Me  (Show Source):
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three consecutive numbers: (x-1), x, (x+1)

((x-1) + x + (x+1))/2 + 2x = 63
3x/2 + 2x = 63
7x/2 = 63
x = 18

The smallest integer is x-1 = 17.

Answer by MathTherapy(10552) About Me  (Show Source):
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One half the sum of three consecutive numbers added to twice the first number is equal to 63, find the smallest integer
Again, that person's answer is WRONG, WRONG, WRONG!! Furthermore, the numbers are not INTEGERS!!

Answer by josgarithmetic(39620) About Me  (Show Source):
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%28%28n-1%29%2Bn%2B%28n%2B1%29%29%2F2%2B2%28n-1%29=63---------literally the problem description. Simplify and solve for n, and find the other two numbers. The middle number is n.






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simplifying,...

%283n%29%2F2%2B2n-2=63
3n%2B4n-4=126
7n=126%2B4
7n=130----------rightside is not a whole-number product of 7.

NO SOLUTION

Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
One half the sum of three consecutive numbers added to twice the first number is equal to 63, find the smallest integer
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Such integer numbers,  satisfying imposed conditions,  DO  NOT  EXIST.

As the problem is worded,  printed,  posted and presented,  it is a  FAKE.


Good for re-cycling,  ONLY.