[note: spoilers for triangle puzzle below]

@silverspawn: Not sure if the integral problem was already sorted out completely, but Kirian is right, you need an constant of integration. So the equation you get is (ln is for barbarians, I use log with base e):

log(y) = K * (-1/2 (t^-2)) + C

(note that you only need a constant on one side of the equation, since if you have constants on both sides, you could just define a new constant as the difference between the two constants you had, and use that instead)

Now you want to use the information y(1) = 5 to eliminate the constant C, and then you'll be able to get a formula

y(t) = *something which depends on t and K*

If all seven rotations are unique, the final answer has to be divisible by 7, doesn't it?

As Tables said, some solutions will be rotated to themselves. However, this is only possible if

*every* rotation of that solution will be identical to it.

[sidenote]You can see this as follows. Suppose that if by rotating 3 times (3/7tau) you get to the same solution, then if you rotate 3*5 times, you'll get to the same solution as well (because you did 3 rotations 5 times, which doesn't change anything), but rotating 15 times is the same as rotating once. And if rotating once doesn't change anything, any rotation doesn't change anything. On a more technical note: the reason is that since 7 is prime, (Z/7Z,+) is a group.[/sidenote]Indeed, there are two solutions which are rotated to themselves, namely:

124 and its rotations (124,235,346,457,561,672,713)

134 and its rotations (134,245,356,467,571,612,723)

And the remaining number of solutions - 133 - is indeed divisible by 7.

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Back to the original puzzle, to see that the answer is 135, consider the following. First note that the problem is equivalent to finding the number of ways to pick 7 triangles such that no two triangles have 2 nodes in common (since a triangle has an edge in common if and only if it has 2 nodes in common).

For convenience we don't have names for the 7 nodes yet. Pick two arbitrary nodes, and call them node 1 and node 2. First let's see how many ways there are to put node 1 in three triangles with the other six nodes. There has to be a triangle with both nodes 1 and 2, and there are

**5** possibilities for that. Call the third node of this triangle node 3, and call the other four nodes 4-7 in any order. For the triangle with both nodes 1 and nodes 4, there are

**3** possibilities (by picking any of nodes 5-7), and for the remaining triangle with 1 as node there's just 1 possibility (picking the remaning two nodes).

Now every node other than node 1 still needs two more triangles. Let's focus on node 2, which still needs triangles with nodes 4,5,6,7. There are

**3** ways to split these four nodes in two groups of two nodes, which is the number of ways to pick the remaining two triangles containing node 2. Also node 3 needs two more triangles with nodes 4,5,6,7, so there are also

**3** possibilities for that.

In total there are 5*3*3*3=135 possible colorings.