Question 13350
we want 25<sum<45.
Now sum = x + (x+2) + (x+4)
        = x+x+2+x+4
        = 3x + 6
we have:
25<3x+6<45 subtract 6 from all sides
25-6<3x<45-6
19<3x<39 divide 3 from all sides
6.333333<x<13
so,
7<=x<13 would be the set of all integers but since we are dealing with evens 8<=x<=12