Question 227190
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Let *[tex \Large x] represent the first odd integer.  Then the second consecutive odd integer has to be 2 more than that (if 5 is the first one, 7 is the next and so on), so *[tex \Large x+2].  The third must be two more, i.e. *[tex \Large (x+2)+2=x+4].

The result of adding the first and the third:  *[tex \Large x + (x + 4) = 2x + 4]

Five times the second integer is *[tex \Large 5(x+2)=5x+10].  87 less than that is *[tex \Large 5x+10-87=5x-77]

So we can write:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x + 4 = 5x - 77]

Now all you need to do is solve for *[tex \Large x], then add 2, then add 2 more and you will have the third integer.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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