Question 760068
In a list of consecutive integers you may find
...., n-3, n-2, n-1, n, n+1, n+2, n+3, ...., in that order.
In a list of consecutive odd (or even) integers you find
...., n-6, n-4, n-2, n, n+2, n+4, n+6, ...., in that order.
So if the third of your consecutive odd integers were {{{n}}},
the first would be {{{n-4}}}, and the second would be {{{n-2}}}.
Three times the first would be {{{3(n-4)}}} and
three more than twice than twice the third number would be {{{2n+3}}}
Then, {{{3(n-4)=2n+3}}} --> {{{3n-12=2n+3}}} --> {{{3n=2n+3+12}}} --> {{{3n-2n=3+12}}} --> {{{highlight(n=15)}}}