SOLUTION: Find three consecutive even integers such that twice he sum of the first and third integers is twenty-one more than the second integer. I know that my integers will be X, X+2, X

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find three consecutive even integers such that twice he sum of the first and third integers is twenty-one more than the second integer. I know that my integers will be X, X+2, X      Log On


   

Question 708125: Find three consecutive even integers such that twice he sum of the first and third integers is twenty-one more than the second integer.
I know that my integers will be X, X+2, X+4. I have set this up 3 or 4 times but when I try to solve I am not getting an answer divisible evenly. So obviously I'm setting it up incorrectly.
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with three even integers, make the first one 2n, where n in any integer. n may be odd or even. Starting with 2n, this is made as an even number.
The next two numbers will be 2n+2 and 2n+4.

Arranging for the other descriptions:
2n+(2n+4) = the sum of the first one and the third one;
2(2n+(2n+4))= TWICE the sum of the first one and the third one;
21+(2n+2) = Twenty one more than the second integer.

Putting everything together according to the phrasing:
2%282n%2B%282n%2B4%29%29=21%2B%282n%2B2%29
.
.
I have worked through this TWICE now, and have obtained n=5/2. This means that the first "even" integer is 2(5/2)=5. This is NOT even.
I believe something is wrong with the problem description.