Question 1180772: There are three consecutive even integers. If twice the first integer added to the third is 268,246 find all three integers.
Found 2 solutions by mananth, greenestamps:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
Let (n-2), n, n+2 be the three even integers.
twice the first integer added to the third is 268,246
2(n-2)+(n+2) =268246
2n-4 +n+2=268246
3n = 268248
n= 89416
The integers are
89414 , 89416,89418
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
While it is often the case in problems with consecutive even integers that using x-2, x, and x+2 for the three integers makes the algebra needed to solve the problem easier, it does not look that way for this problem.
A quick mental solution could go like this:
The third integer is 4 more than the first; so if the sum of the third and twice the first is 268,246 then three times the first is 268,246-4=268,242. So the smallest of the integers is 268,242/3 = 89414.
ANSWERS: 89,414; 89,416; and 89,418
The same solution using formal algebra would look like this:
2(x)+(x+4) = 268246
3x+4 = 268246
3x=268242
x=268242/3 = 89414
89414, 89416, and 89418
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