Three consecutive odd integers have a sum of 45 in algebraci expression?

Answer 1

If x is the smallest odd integer, then x = 2n + 1 for some integer n.

Then the next two odd integers are 2n + 3 and 2n + 5

So the question then becomes:

2n+1 + (2n+3) + (2n+5) = 45

or 6n + 9 = 45 to be solved for n and thence the smallest of the three consecutive odd integers.