Question 661245
Let x be the smallest of the three consecutive odd integers.  Then the next two odd integers are x+2 and x+4.  If these are the sides of a triangle with perimeter 51, then we have:
{{{x + (x+2) + (x+4) = 51}}}
{{{3x + 6 = 51}}}
{{{3x = 45}}}
{{{x = 15}}}
So the three sides are 15, 17, and 19.