Question 659725
Call the first integer x.  Each consecutive odd integer is two more than the last, so the four are x, x+2, x+4, x+6.  Their sum is 432.<P>
x+x+2+x+4+x+6 = 432<P>
4x+12 = 432<P>
4x = 420<P>
x = 105<P>
The integers are 105, 107, 109, 111
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