Problem Solving

How many triangles can be formed so that the length of the sides are three consecutive odd integer and perimeter less than 1000.

Md.Nazrul Islam wrote:How many triangles can be formed so that the length of the sides are three consecutive odd integer and perimeter less than 1000.

Let the 3 sides be x, x+2 and x+4.

The length of the longest side must be less than the sum of the lengths of the other two sides:
x+4 < x + x+2
2 < x.
Thus, the smallest possible value of x is 3, implying a triangle with sides of 3, 5 and 7.

The perimeter must be less than 1000:
x + x+2 + x+4 < 1000
3x < 994
x < 331.33.
Thus, the greatest possible value of x is 331, implying a triangle with sides of 331, 333 and 335.

Thus, the shortest side can be any odd integer between 3 and 331, inclusive.
To count evenly spaced integers:
Number of integers = (biggest-smallest)/interval + 1, where the interval is the distance between one integer and the next.
The interval between consecutive odd integers is 2.
Thus, between 3 and 331, inclusive, the number of odd integers = (331-3)/2 + 1 = 165.