brain teaser again.................?

Question

if u r given 'n' rods of length{1,2,3,4.....n} respectively.
what is the maximum no. of triangles that can be formed using these rods at a time?
one rod can be used only once.........
for example if n=5 , the max. no. of triangles formed will be 1..........using the combination(3,4,5)

so once again read it carefully....
you cant use one rod more than once
suppose n=5 ,then the answer is 1
using combination(3,4,5) or (2,3,4)
if u r using rod "3" in one combination
you cant use it in another combination..
in other words you have to show all your combinations keeping them on a table at a time....and you have to exploit maximum no. of rods possible
for simplicity try to device a method for n=100 and generalise it for n.....
hope you all understand the question better now.........

friends...be serious about this problem..
dont just count your points......

Answer 1

The maximum has to be a number less than n/3.
Clearly length 1 can never be used.
Other than this, any 3 consecutive rods can form a triangle, because
k + (k+1) > (k+2) as long as k > 1

So drop #1, and then take the rods 3 at a time until you have less than 3 left at the end.
Generally it will be 1/3 of the highest multiple of 3 less than or equal to n-1.

[(n-1) - mod(n-1, 3)] / 3

Answer 2

If n is divisible by 3 then total triangle =n/3.
If n is not exactly multiple of 3 then triangles will be either (n-1 )/3 or (n-2 )/3 depending upon the remainder of division by 3.
For example if n=100, then triangles =100/3 or 33 triangles

so the generalisation can be :
total number of triangles = n/3 where nis exactly divisible by 3

Answer 3

if u hav 'n' no. of rods n if u want 2 find d no. of triangles that can be formed then u hav 2 use d theory of relativity with d newtonian laws i.e if u divide the biggest integral of them wid the smallest side then only u will have the perimeter and once u hav d perimeter u hav 2 form three equations 2 find the sides or even u can go reverse frm here...u understood naa

Answer 4

max number of triangles formed with such n rods is
= m, where (n -- 1) = 3m + k, k ≤ 2.

Answer 5

The rod with length of one is not usefull at all. because one rod can be used only once, the other triangles can be somthing like (2,3,4) & (5,6,7) and so on. Therefore the number of triangles is the absolutevalue of (n-1)/3.

Answer 6

n x(n-1) x( n-2)

Answer 7

1 If n is divisible by 3 with remainder Zero, then n/3 triagles
2 If n is not divisible by three, then divide n in to two parts a and b.
3 Where a is divisible by 3 and b is ether 1 or 2
4 Then no of triangles are a/3 with either one or two rodes are in spare

Answer 8

wow, well blown away on this one. lol

Answer 9

hummmnnn,,,,uh??wat?