Leaderboard

Assume that the earth is a perfect sphere with a circumference of 40Mm. Santa needs to travel from the North Pole to the South Pole while avoiding daylight. Assuming that he can go faster than the speed of light, what path would be the best path to take and what is the slowest he can travel?

So let's say I have a time machine, and I put £1,000 into a bank account (at time t=0). Now, if I don't touch it for a year (call this t=2), I'll have some interest on my original £1,000. Let's say the interest rate is 1% per year (works for any interest rate really). Then at t=2, my balance is £1,010. So if I ask to withdraw all my money, so I now have £1,010 in my wallet. Then I use my time machine to travel to the day after I deposited the original £1,000, call this t=1,and deposit half the money in my wallet, that is, £505. I then go back to t=3, where I still have the other £505 in my wallet. BUT... If I deposited the money, then at t=1 I had £1,505, instead of £1,000. So at t=2, adding 1% interest gives me £1,520.05, so THAT'S actually how much I withdraw. So I deposited half of that = £560.02 and kept £560.02 for myself at t=3 (I'll give the extra penny to charity) But that's more money, which means it actually become more, and more, and more. Does it converge, or do I have a truly infinite supply? Can we increase our final outcome if the initial parameters change? Let's go to the maths! Let M be the amount of money I have (since all my money at any given point is either in the bank or in my wallet, I only need one letter to denote this. M0 is my starting money at t=0, M1 is money at t=1, and so on. M0 = £1000 Let I be the interest rate, aka 0.01. Let P be the proportion of my money I deposit at t=1 (the rest I keep at t=3, giving the remainder to charity), this = 0.5. Equations: A. M1 = M0+P*M2 B. M2 = M1* 1 + I) C. M3 = (1-P)*M2 Substituting A into B gives M2 = (M0+P*M2)*(1+I) M2/(1+I) = M0+P*M2 M2/(1+I)-P*M2 = M0 M2*(1/(1+I)) - M2*P = M0 M2 * (1/(1+I)-P) = M0 M2 = M0 / (1/(1+I) - P) So the amount we end up with at then end (M3) is 1-P ------------- * M0 1/(1+I) - P If 1-P > 1/(1+I)-P, we made a profit. 1 + I > 1 -> 1 / (1+I) < 1 -> 1 / (1+I) - P < 1-P We made a profit. How much profit? If it's more than I, this was worth it. Let's plug in our values: (1 - 0.5) / (1/1.01 - 0.5) = 1.0202020202... ​We (slightly more than) doubled the interest rate. In fact, changing the interest rate with P = 0.5 doubles the interest rate What about different values of P (putting depositing more at t=1)? Lower values of P end up giving results close to 1.01, but higher values give more profit: for example, (1 - 0.9) / (1 / 1.01 - 0.9) 1.11, effectively bumping up the interest rate to 11%. However, going too high gives negatives, so what's the highest point we can go to? Answer: P = 1/1.01 = 100/101. Then our money goes to infinity. (as we divide by 0) So for interest rate I, depositing (1/(1+I)) of the money at t=1 when in the past means you end up with an infinite amount. Of course, due to rounding, you can't deposit exactly this amount, but you get more money each "iteration" so you can round it more accurately each time. And if you give the remainder of division after rounding to charity, you'd be helping infinitely too. (in most cases) The only limit is the amount of money your wallet can carry at once as you travel through time.