# Deal or no deal?

You wouldn’t think that scientists get to travel very much, but so far I have visited every continent on Earth but one: Alaska (hmm…) Yet, even before I was a world-renowned Professor at the top university in the universe (or, as I tell my parents “postdoc at Caltech”), I had penpals (when I was half my age – my age is a power of two – the concept of penpal was still alive and strong) from places like Argentina, Cyprus, Germany and Romania (gotta love international math and sports competitions). The friends I made were often local kids that would hang out with the visiting athletes (or mathletes, depending on the nature of the competition), so the reaction I got whenever I mentioned that “Μου αρέσει το volleyball και ο στίβος, αλλά τρελαίνομαι για τα μαθηματικά!” (I love volleyball and track & field, but I am crazy about math!) was pretty uniform: “Eh?”

You and I know why math is so beautiful, but in case someone else is reading, I would like to share the source of that excitement with problems that seem impossible to solve, until they don’t. Math and sports are the same in some sense: In order to have a chance to succeed, you must overcome yourself first. What does it mean to overcome one’s self? It is actually pretty simple:

Problem 1: If $1+2+4+8=15$ and $1+2+4+8+16=31$, what is $1+2+4+8+16+32+64+128+256+512+1024$?

If you haven’t seen a geometric series before (the sum of numbers where the next number you add is a multiple of the previous one; in this case it is twice the previous one), then you will probably throw your hands up in the air and declare defeat. For three reasons:

1. You don’t have a calculator,
2. You don’t know how to add without a calculator, and
3. You don’t care!

So, what do you do? Well, the next step depends on your answer to this question: Are you curious to know what the answer to Problem 1 is? No. OK, fair enough. Here it is anyway: 2047. I guess you were right; it is a pretty random-looking number. But why is it 2047? Look at the last number in the sum: 1024. If we multiply this number by 2 we get 2048…!
Same pattern appears for $2 \times 8 = 16 =15+1$ and $2 \times 16 = 32=31 +1$. The answer seems to be one less than twice the last number in the series. Why?

Everyone starts from humble beginnings. Even Pierre de Fermat.

Problem 2: Prove that $1+2+4+\ldots + 2^n = 2^{n+1}-1$.

Whoa! Prove? What do you mean prove? Didn’t you just show that it works for three randomly chosen sequences of numbers? Doesn’t that prove that it always works?

No.

Let’s go back to overcoming one’s self. You have figured out what the answer to Problem 1 is, so the answer to Problem 2 seems obvious. It is a gut feeling, an ancient instinct that you are always going to get the right answer if you double the last number and subtract one, right? So, here is a question for you:

Uhmmm, let me see… The cube root of 3,141,592… Oh great, now I am melting.

Problem 3: I will give you $333,333 for all the M&Ms you put on a 4-by-4 wooden chessboard of your choice, but you have to put 3 times as many delicious M&Ms in each consecutive square of the board starting with one M&M – on average, you can sell 1Kg of M&Ms for$5 and each M&M weights about 1g (the things you learn here). Oh, and I get to keep the useless chessboard.

Deal or no deal?

This entry was posted in Real science, The expert's corner by spiros. Bookmark the permalink.

## About spiros

Spyridon Michalakis does research on quantum many-body physics and topological quantum order at Caltech's Institute for Quantum Information and Matter, where he is also the manager for outreach activities.

## 22 thoughts on “Deal or no deal?”

1. If I am correct, you have a deal.
So, if you did not signed a really nice contract with M&M for the commercial, you’ll pay a lot of money for a wooden chessboard – around 225 thousand $. But it will be quite huge chessboard 🙂 (just kidding, I love M&M ;)) Cheers 2. No deal. Although the chessboard will have 64570081 M&Ms, worth about$300,000 (rounding down from $322850 since M&M value and weight is only given in one significant digit), the weight of the M&M is over 60 metric tons. I would essentially be making a few thousand dollars for shipping and labor of giving you 60+ metric tons of ‘stuff’, which is highly discounted from what UPS charges. • Excellent answer, especially the shipping bit, but 3^0+3^1+3+3^2+…+3^n. Is n=15 or 16? • As the chessboard is 4×4, it has 16 squares to put the M&Ms in. You start from 1, then 3, etc., which is 1+3+3^2+…so the last square will have 3^15 M&Ms. • Correct. I am happy to see that the post generated such great responses. I am still hoping for someone to do some research on how cheaply one could create a wooden chessboard the size of a few football fields. And no comments about Alaska yet (though plenty privately). Everyone is so polite… or they got the joke? • My calculation says the chessboard will cost around$210.000! If we try to save some money, maybe we could make it for approx. $150.000. Together with all the M&Ms, which we need around 21 tons that cost around$107.615, the final bill is 317.615 (or $257.615 if we manage to save money). And this does not include any discount that we might get for buying a lot of material 🙂 If you are interested, I can write the estimation in details. Cheers P.S. As for the Alaska, I did note but I forgot about it once I start to think about the problem. I suppose I am polite then. • I would be very interested to see the calculation for the chessboard. Including the type of wood you would use! 3. Of course (I decided to make it in new thread, so it is easier to read). So, about the chessboard…The biggest square in the chessboard will contain 3^15=14.398.907 M&Ms. I assume that one M&M can fit a square with size 1x1cm. The square root is then approx. 3787, so that means the square of size 3787×3787 cm will do the job. This is around 38 meters, so the size of the whole chessboard would be around 152×152 meters. I suppose the plywood would do the job as a cover. The price of m^2 of 1cm tick plywood is around$9, so we have to pay 152m^2×9 which is $1.369. The price is for birch or poplar (pine is also of similar price, a little less). Then, I also suppose we cannot put it on the plain ground. So we have to buy some joists. The price is$550 per m^3 for spruce joists, but how many we need so that we have a stable chessboard?
IMHO, one has to be 152m long, and I suppose that it can be 10cm wide and also 10cm thick. So, one joist is around 152 x 0.1 x 0.1 which is 1.52 m^3.

I assume it will be enough to put one joist per meter, so we will need 153 joists. This means 153 x 1.52 x 550 which is around $127.908. Next, we need connect it, so we will have to put some$1.5 nails. If we put 305 nails on every joist (a nail per 1/2m), this will cost us around $70.000. Furthermore, we probably have to find a land to put it and make it more or less flat. I assume it could cost$10.000 (the work, not the land; really a rough estimation).
Finally, we need to paint it. A can of 1L of paint can paint 16m^2 according to the specification, and it costs $15. This means we need$150 for paint.

The final bill for the chessboard is then $209.426! Together with all the M&Ms, which we need around 21 tons that cost around$107.615, the final bill is 317.041!

Now, maybe we can try to save some money. Joists are the most expensive, so we can try to take only 5cm tick and 5cm wide joists, and pay only $63.954. The final bill is then$253.087. Or maybe we could use something else, or try to optimize and use as less joists as possible. But I did not think too much about it.

Maybe, we can also use glue instead of nails (but I did not research the cost and amount of resources).

Or maybe, we can ask for some discounts on all this 🙂

So, that’s it. If anyone think I made a mistake in the calculation or the estimation, I would sincerely like to hear it.

Cheers

P.S. The last sentence in my previous comment should have a question mark and a smiley at the end, I did not want to sound affirmative about the politeness, it was a joke.

• Excellent job Filip. Now the question is: Should I pay $333,333 for all these M&Ms and the chessboard? • Well, I think you should not. Even if you could sell all the M&Ms and chessboard material for the same price that I bought it, you will lose some of your money (and if I save and buy those 5cm joists, you will lose quite a lot). • How much do you think I could charge families to come look at the world’s largest chessboard ever created? • Hmmm…depends where we put the chessboard, maybe$5-10.
You will need to have a lot of visits then in order to return the money, lets say around 15.000-20.000 (roughly estimation, still depends on the price you charge, the tax you pay, the joists I choose and some other details that we neglected). The bad thing is that you will need some time to get the money back. The good thing is that, if people are interested, you will continue to make money. So, at the and maybe we both get a good deal 😉

• Exactly. I think within a year I would recoup most of the money and you would get a pretty nice deal too. Even if you neglected to include all the labor costs in your calculations. I assume you have a good network of carpenters on speed-dial 🙂

• Actually I did not. You did not specify place and time constraints for the chessboard. I would build it in my small city (Serbia), where usually we do all the work for ourselves, so I will count on the help of my father (who would do it for free, anyway he´s doing some similar carpenter work from time to time and has experience). I think the two of us could do it.
The problem is that I assumed that all the material comes in huge size (e.g., 152x152m plywood, 152m long joists, which is not realistic). In reality, it will be smaller so we will just need to spend extra money on material to connect those – more nails – but in any case at the end we both get a nice deal I think 🙂

• An entrepreneurial scientist with woodworking experience! I almost think we should go ahead with the deal. If only I had $333,333 to spare on M&Ms and the largest wooden chessboard in the world… 4. spiros wrote (September 3, 2012): > Problem 3: > I will give you$333,333 for all the M&Ms you put on a 4-by-4 wooden chessboard of your choice, but you have to put 3 times as many delicious M&Ms in each consecutive square of the board starting with one M&M

Is that:
3 times as manyas had been put on the previous field ?,
or
3 times as manyas had been put on the entire board ?

• The first. But please solve the problem if the second statement is what I meant. I would like to see how many M&Ms that would be.

5. spiros wrote (September 6, 2012 at 5:34 pm):
> […] I would like to see how many M&Ms that would be.

Right …

And I would like to see a comment preview implemented here, fer crying out loud.

Deal ?

• Sorry Frank. I only recently learned how to use CSS to center latex with the baseline of the text. I will look into the preview feature, but it may be that WordPress doesn’t give me enough flexibility to implement it. In the meantime, if something works inside the math environment of latex, then it should work in the WordPress latex environment. Outside of commands in the math environment it will not work.

6. You know that moment when you realize you probably made a mistake, but you did not even think about the problem? Well I had it this morning while on my way to the lab.
I made quite serious mistake (shame on me) in calculating the plywood resources. I calculated only for the 152m^2, and not for the whole board. This drastically increases the expenses, so much that actually I lose and you win at the very beginning!