Question 1071835: The sum of the squares of six consecutive positive even integers is 8284. What are the six numbers?
Found 3 solutions by stanbon, MathTherapy, greenestamps:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The sum of the squares of six consecutive positive even integers is 8284. What are the six numbers?
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1st:: 2x-4
2nd:: 2x-2
3rd:: 2x
4th:: 2x+2
5th:: 2x+4
6th:: 2x+6
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Equation:
sum of squares of those six numbers = 8284
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I'll leave that to you.
Cheers,
Stan H.
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Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
The sum of the squares of six consecutive positive even integers is 8284. What are the six numbers?
Let the 3rd integer be T
Then 2nd and 1st are: T - 2, and T - 4, respectively
Also, 4th, 5th, and 6th are: T + 2, T + 4, and T + 6, respectively
We then get: 
 ------ FOILing binomials
(T + 38)(T - 36) = 0
T + 38 = 0 OR T - 36 = 0
3rd integer, or T = - 38, or 36
With 3rd integer = - 38,
2nd integer = - 38 - 2, or - 40
1st integer = - 38 - 4, or - 42
4th integer = - 38 + 2, or - 36
5th integer = - 38 + 4, or - 34
6th integer = - 38 + 6, or - 32
With 3rd integer = 36,
2nd integer = 36 - 2, or 34
1st integer = 36 - 4, or 32
4th integer = 36 + 2, or 38
5th integer = 36 + 4, or 40
6th integer = 36 + 6, or 42
Therefore, integers are:  OR 
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The answers posted so far both let one of the 6 numbers be x, or 2x. That leads to solving a quadratic equation that involves an x term. The algebra is easier if you let the two "middle" numbers be x-1 and x+1; that way, when you add the expressions for the squares of the 6 numbers, the x terms all cancel out, making the solution much easier.
So let the 6 numbers be x-5, x-3, x-1, x+1, x+3, and x+5. Then the sum of the squares is
So then
And so the 6 numbers are 32, 34, 36, 38, 40, and 42.
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