Question 209999
Let x = first odd integer
x+ 2 = second odd integer
x+ 4 = third odd integer


The sum of these integers (add them up!) will be 3x + 6.


The inequality will be 
64<3x+6 <74


Subtract 6 from each part of the inequality:
64-6<3x+6-6<74-6
58<3x<68


Divide by 3:

{{{(58/3)<x<(68/3)}}}

19.333. . . < x < 22.666. . .


Since these are INTEGERS you are looking for, the only ones that fit this description would be x= 20, 21, and 22.   You must reject the x=20 and x=22 since these are NOT odd integers.  The only solution would be  


x=21 
x+2 = 23
x+4=25


The sum is 69.


R^2


Dr. Robert J. Rapalje
Seminole Community College