Question 548494
Let the lesser of the three consecutive odd integers be {{{n}}}.
Consecutive odd integers differ by 2.
The other two consecutive odd integers would be {{{n+2}}}, and {{{n+4}}}.
Do not worry about their being odd at this point. We can always discard any even solutions at the end. (And, by the way, we would set the problem the same way if it was about three consecutive even numbers).
The sum of the first two is {{{n+(n+2)}}}.
29 more than the third is {{{(n+4)+29}}}
The problem says that those two expressions have the same value, so we write
{{{n+(n+2)=(n+4)+29}}}
Then we simplify those expressions
{{{n+(n+2)=(n+4)+29}}} <--> {{{n+n+2=n+4+29}}} <--> {{{n+n+2=n+33}}} <--> {{{2n+2=n+33}}}
Then we subtract 2 from both sides
{{{n+n+2=n+33}}} <--> {{{n+n+2-2=n+33-2}}} <--> {{{n+n=n+31}}}
Then we subtract n from both sides
{{{n+n=n+31}}} <--> {{{n+n-n=n+31-n}}} <--> {{{n=31}}}
So our first odd integer is 32. The others are 33 and 35.
Let's verify. The sun of the first 2 is {{{31+33=64}}}
and 29 more than the third one is {{{35+29=64}}}.
We found the three consecutive odd integers the problem talks about. They are
31, 33, and 35.