In a Dutch High School, students received classes about Bitcoin, and have to answered some hard questions. Some crypto enthusiast does not know how to answer, here are the link to the news,
https://coinidol.com/exam-on-bitcoin-math/, and I will quote the questions and the answers from this link
https://www.reddit.com/r/CryptoCurrency/comments/8jd9hu/this_is_fucking_awesome_hahahaha_about_200k_dutch/dyz54ni/
Take the test and see you answer The question is as follows, don't mind my subpar translating skills.
Bitcoin is a digital currency that only exists online. It has existed since January 1st 2009 and can be used as payment method in webstores and for other online services. Bitcoin is not, like standard money, made by a central bank. Instead, all bitcoin that exist are created by having computers participate in solving specific mathematical problems. This works as follows: everyone can run special software on his or her computer that participates in solving such a mathematical problem. The owner of the computer that solves the problem receives 25 (newly created) bitcoin as a reward. Because it was the case that in 2014 such a problem is solved every 10 minutes, 25 new bitcoins were created every 10 minutes. On January 1st there were (approximately) 12.2 million bitcoin.
Question A: Assuming the above, calculate in what year the amount of bitcoin exceeded 18 million, given that the speed at which they are introduced does not change.
Solution: Each day there are 60 / 10 * 24 * 25 = 3600 bitcoin to be earned. Given that there is 18-12.2 = 5.8 million bitcoin left to be earned, that's 5.8 * 106 /3600 = 1611 days, bringing us from 2014 to 2018.
In reality the rate at which bitcoin are introduced does not stay equal to 25 bitcoins every 10 minutes. This rate reduces. During the first four years, from January 2009 to January 2013, the reward was equal to 50 bitcoin. The reward gets reduced by half every four years: from January 1st 2013 to January 1st 2017 the reward is 25 bitcoin, for the next four years it is 12.5, and so forth.
Question B: calculate from which year on the reward will be less than one bitcoin.
Solution: the amount of bitcoin y given as a reward is given by y = 50 * 0.5x where x is the amount of 4 year periods that have passed since January 1st 2009 (the first period is given by x = 0). So we need to solve 50 * 0.5x = 1. In Mathematics A you're allowed to solve this directly with your calculator, else just divide by 50 and take the log. This gives that x = 5.6, so it will be after 6 periods. 6*4 + 2009 = 2033.
The total amount of bitcoin that can be put into circulation is limited. This is partially due to the reward halving every four years. The amount of bitcoin that is currently in circulation can be approximated by the following equation: C = 21 - 21 * 0.50.25*t Here C is the amount of bitcoin in millions and t equals the amount of years with January first 2009 given by t = 0.
Question C: Determine the maximal amount of bitcoin that can be in circulation based on the above equation based on reasoning.Solution: 0.50.25*t goes to 0 for large t, so that C = 21 - 21*0 for large t. The limit is therefore 21 million bitcoin.
To regulate the total amount of bitcoin in circulation, not only the number of bitcoin rewarded per solution is reduced, but the difficulty of the mathematical problems being solved is also increased. That is because more and more people use their computer to solve the problems. The difficulty of the problems being solved scales exponentially according to the equation
D = 3.65 * e0.533t
Here D is the difficulty of solving the problem and t is the time in months, with T = 0 on January 1st 2013. Larger D implies that the problem is harder to solve.
Question D: Write down the equation for the derivative of D and argue that D is an increasing function, that increases by an increasing rate.Solution: The derivative is 0.533 * 3.65 * e0.533t by virtue of how the derivative of ex works plus the so called chain rule; if f(x) = ex then f'(x) is ex, and if g(x) = eh(x) then g'(x) = h'(x)eh(x). Now, e^(0.533t) is > 1 for t > 0 so the derivative is positive and D is an increasing function, and e0.533*t2 > e0.533*t1 for t2 > t1 > 0 so that the derivative grows with time, leading to an increasing rate at which D increases.
The above formula for D can be rewritten in such a way that you can find the amount of months needed to reach a specific D.
Question E: Rewrite the above equation for D in such a way.Solution: D = 3.65 * e0.533t, taking the log (= base e logarithm, they call it ln in Dutch high school math but I dont believe this is universally true) on both sides we get that log(D) = Log(3.65e0.533t) = Log(3.65) + Log(e0.533t) = Log(3.65) + 0.533tLog(e) = Log(3.65) + 0.533t so that we can write down that t = Log(D/3.65)/0.533 using the rules of the Log function (Log(xy) = Log(x) + Log(y), Log(xy) = ylog(x))
