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?