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

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)   (Show Source): You can put this solution on YOUR website!
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)   (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

Again, that person's answer is WRONG, WRONG, WRONG!! Furthermore, the numbers are not INTEGERS!!
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
---------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,...

----------rightside is not a whole-number product of 7.

NO SOLUTION
Answer by ikleyn(52797)   (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
~~~~~~~~~~~~~~

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.

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