Question 1180329: What are three consecutive even integers if the sum of their squares is 1208?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
The integers are (n-2), n and (n+2).
The equation to find "n" is
(n-2)^2 + n^2 + (n+2)^2 = 1208
n^2 - 2n + 4 + n^2 + n^2 + 2n + 4 = 1208
3n^2 = 1208 - 4 - 4 = 1200
n^2 = 1200/3 = 400
n =  = +/- 20.
ANSWER. There are two such sequences.
One sequence is 18, 20, 22. Another sequence is -22, -20, -18 (in ascending order).
Solved, answered and explained.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The good solution from tutor @ikleyn is a good example of how choosing the 3 consecutive even integers as x-2, x, and x+2 can make a problem much easier to solve using formal algebra than calling them x, x+2, and x+4.
Keep that in mind when you solve other problems about consecutive integers.
The same concept can be used to solve the problem informally in a very short time.
The three integers are close together, so the sum of the squares of the three integers is close to 3 times the square of the middle one.
1208 divided by 3 is a bit more than 400, and 400 is 20^2.
So the numbers should be 18, 20, and 22. Perform quick calculations to verify that 18^2+20^2+22^2=1208.
Then remember that since the problem is about the sum of the squares of the three integers, another solution is -18, -20, and -22.
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