Question

How many triangles can be constructed so that length of its sides are three consecutive odd integers and its perimeter is less than 1000.

Answer 1

Let the 3 sides of the triangle be $2n+1,2n+3,2n+5$, where $n\ge 0$. So the triangles will be $(1,3,5)$for $n=0$ and $(3,5,7)$for $n=1$ and so on.

A condition the sides of a triangle should satisfy is the Triangle inequality. Sum of any two sides should be greater than the third side.

$\Rightarrow 2n+3+2n+5>2n+1$

$\Rightarrow 2n+8>1$

$\Rightarrow 2n>-7$

$\therefore n\ge 1$

Since, the perimeter is less than $1000$ units.

$2n+1+2n+3+2n+5<1000$

$6n+9<1000$

$6n<991$

$n<165.17\approx 165$

Hence, this is the answer.