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Posted on Jan 02, 2017
Let's say your 3 numbers are a, b and c. Because they are consecutive and odd, you can write them like this:
a = 2x-1, b = 2x+1, c = 2x+3
Now, your problem looks like this:
2(2x-1)(2x+1) = (2x+1)(2x+3) + 7
2(4x^2-1) = 4x^2+6x+2x+3 + 7
8x^2-2 = 4x^2+8x+10
4x^2-8x-12 = 0
4(x^2-2x-3) = 0
x^2-2x-3 = 0
The solutions for this quadratic equation ( see more about this kind of equations here: http://en.wikipedia.org/wiki/Quadratic_equation ) are x=3 or x=-1. So, you've got two sets of solutions:
 a = 5, b = 7, c = 9
 a = -3, b = -1, c = 1
Posted on Oct 19, 2009
Let X be the lowest of the 3 integers
The sum of the 3 integers can be represented as X + (X+1) + (X+2).
Set this sum equal to 378 and solve for X.
X + (X+1) + (X+2) = 378
3X + 3 = 378
3X = 375
X = 125
Since X is the lowest of the 3 integers the other 2 will be X+1 and X+2 or 126 and 127.
Therefore, the answer is 125, 126 and 127
Posted on Jan 28, 2011
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