Question
I'm going to be attending something called SketchKon - a conference being held in Pasadena, California this year run by an online art school called Sketchbook Skool. 500 people are expected to attend (aside from presenters, instructors, and the like). Many people are making stickers to trade with other attendees. If all the attendees make enough stickers/ATC to trade with everyone else at least once, how many stickers will be at the conference? If everyone manages to trade with everyone else at least once, how many trades will occur over the course of the event?
Answer
at least 249,500 stickers; at least 124,750 trades
Analysis
This question is a wonderful opportunity for me to present things in a slightly different way than I usually do, by which I mean that I'll be writing directly to artists. That opens up an approach to math that I don't normally get!
And just to make things easy, I'll refer to the items that are going to be traded, whether stickers or Artist Trading Cards, or whatever, as "stickers".
I'll answer the "number of stickers" question first and then tackle the number of unique trades second - and it's there that I'll play with art and math together.
Number of stickers at the event
If each person makes enough stickers that they can trade with each other person at least once, this means that each person will be bringing at least 499 stickers (if I were one of the 500 people attending the event, there'd be 499 other people there, and so I'd need 499 stickers to trade with other people).
With 500 people, each with 499 stickers, we multiply the two numbers together to get:
500 X 499 = 249,500 stickers at the event.
Just for fun, let's assume that each sticker is roughly 0.5 grams (or roughly 0.02 ounces) - which is roughly the weight of an A7 sized piece of 80 GSM paper. The weight of all those stickers would be:
249,500 X 0.5 = 124,750 grams, or approaching 125 kilograms (for those conversant with pounds, it approaches 275 pounds).
Remember - I've assumed that each person brings only one sticker for every other person. There could be more!!!
Number of trades at the event
Here's the fun question - I'm going to use two shapes, two equilateral triangles (triangles with equal sides), to figure out the number of trades!
Here's how I'll do it - I'm going to assume for a moment that the sticker trade will occur in a very unartistic and non-spontaneous way - I'm going to have participants, one-by-one, enter what I'll call the Sticker Exchange Room. This will make sure that everyone gets to trade with everyone else in an organized manner (important for the mathy-ness of the question but totally not fun!)
The first person enters the room. Let's call this person A.
A
A is alone and very sad so let's not leave A alone for long. Let's have B come in:
B
A
A and B trade a sticker. 1 trade made! Now we can let in C. We'll have A and B stand next to each other while C trades with them:
C
A B
C trades with A and B: 2 trades. And if we include the trade that A and B did already, 3 trades in total have happened. Now let's let in the next person, D:
D
A B C
D trades with A, B, and C: 3 trades. And if we include the 3 trades that have already happened, 6 trades in total have occurred.
At this point, you might be wondering where the first triangle is. Well... we've begun building it. If we look at the numbers 1, 3, and 6, and we arrange that number of items (such as trades of stickers) into a triangular pattern, we get those numbers:
1 = 1 + 0 = 1 trades involving 2 people
1 1 = 2 + 1 = 3 trades involving 3 people
1 1 1 = 3 + 3 = 6 trades involving 4 people
and we can continue to build this triangle as more and more people enter the trade room. In fact, we can do this all the way to 500 people! But this would be a very long way to figure out the number of trades.
Let's find another way. To do that, we can use another triangle, this time called Pascal's Triangle. The very top bit of it looks like this (we can continue adding lines onto it into infinity):
Each number in the triangle is found by adding the two numbers above it. For instance, in the third row, the 1 2 1 row, the 2 is found by adding the 1's above (and slightly to either side) of it. If there is no number present (say like with all those 1s running down either side), the black space is a 0.
(Gif by
Hersfold on the English Wikipedia -
Own work, Public Domain,
Link)
If you look closely, you'll see, starting at the third 1 down on either side and then running diagonally, those same "
Triangular Numbers" we were finding when we were talking about individual trades: 1, 3, 6, 10 and then further down 15 and 21 (15 being the number of trades involving 5 people and 21 trades coming from 6 people). All we have to do is extend this triangle down 500 places and then we can find our number of trades!
But that would be a very big picture. And so at this point I'm going to use another property of Pascal's Triangle (the ability to use a calculation called a
Combination), which will allow me to find any value on it without doing a lot of writing. I'll cut to the chase so that the math-squemish need not close their eyes tightly! We'll use the combination
:
And just for fun - if we had the Sticker Trading Room and we were able to manage one trade every second, it would take roughly 35 hours to do the trading.
So... ignore my idea of a Sticker Trading Room!
Have fun at the conference!
~~~~~
Questions and comments always welcome!
