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
- Senior | Next Rank: 100 Posts
- Posts: 32
Let the 3 sides be x, x+2 and x+4.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.
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.
