Question 909555
Consecutive integers are integers that follow in sequence, each number being {{{1}}} more than the previous number, represented by {{{n}}}, {{{n +1}}}, {{{n + 2}}}, {{{n + 3}}}, ..., where {{{n}}} is any integer. If we start with an {{{odd}}} number and each number in the sequence is {{{2}}} more than the previous number then we will get {{{consecutive}}}{{{ odd}}}{{{ integers}}}.

if the lengths of the sides of a triangle are consecutive odd integers, then they are {{{n}}}, {{{n+2}}}, {{{n+4}}} 

if the perimeter is {{{39m}}}, then  {{{n+(n+2)+(n+4)=39m}}} ...solve for {{{n}}}

{{{n+n+2+n+4=39}}}

{{{3n+6=39m}}}

{{{3n=39-6}}}

{{{3n=33}}}

{{{n=33/3}}}

{{{highlight(n=11)}}}

now find sides {{{n+2}}} and {{{n+4}}} 
{{{n+2=11+2=highlight(13)}}}
 {{{n+4=11+4=highlight(15)}}} 


 so, the measure of each side is: {{{highlight(11m)}}},{{{highlight(13m)}}}, and {{{highlight(15m)}}}