Woah! Chaz, is that some kind of "new math"? Triangle has 3 sides. Each side 17 widgets long. 3 x 17 = 153? Help!

You're forgetting the interior of the triangle.

Look at 10 bowling pins. Four on a side — 1-2-4-7, 1-3-6-10, 7-8-9-10 — but there's still 10 pins. And we never mentioned the 5-pin even once.

(I once converted a 6-7-10 split, which used up rather a large percentage of my allotment of divine interventions.)

You know, I've been watching you post links to CotV pretty much since #1, and keep waiting for you to come up with a blank on the numbers, but I'm thinking now that it just won't happen.

My hat is off to you, sir.

Well, originally I did riffs on the Carnival itself, or on the format used, or on the name of the blog hosting it, but eventually I got to the point where the numbers themselves became something to play with.

Besides, in my younger days as a math geek, I once heard from a teacher that "there are no uninteresting integers." Now's my chance to see for myself.

I once converted one too. Too bad it was only in practice and not recorded anywhere!
I was actually just going for the 6-10 but I managed to hit it just so that the 6 flew across and took out the 7. It was gorgeous to watch and I have not been able to re-create it :(

I was actually in a league when I pulled mine off (late 1974); it seriously jeopardized my chances for the Least Improved trophy.

Triangular number can be expressed thusly, where n is the number of items making up the side of the equilateral triangle:

n + 1 x n/2

In the case of the pins in bowling, n is 4, so the equation is expressed, essentially, as 5 x 2.

Where n is an odd number, we can adjust the equation by leaving the division by 2 for last:

(n + 1 x n)/2

Therefore we get (17 x 18)/2, which according to my calculator works out to (306)/2, or 153.

Q.E.D.