Giant numbers (Ex: how many primes are smaller than the largest known?)

If two 'opponents' at Go are actually collaborating to make the game as long as possible, how many turns might the game last (on a 19x19 board)? (???) This is a great thought question. The starting point, though, is to ask whether there are cycles possible with the rules of go, in which case the number of turns is not finite. It may not be immediately obvious if you don't know the game well, because pieces don't move, but there are http://senseis.xmp.net/?RepeatingPositions to construct a cycle using the basic stone placement rules. The simplest of these is just http://senseis.xmp.net/?Ko, which is prohibited by all rule sets. The reason to prohibit it is that even in an adversarial situation, it might be optimal play to not "give in" -- breaking the cycle means you lose a piece, and so optimal play means a non-terminating game. The basic ko rule in particular prohibits recapturing a piece right after it has captured a ko stone. So then, you might ask, what if there are two kos going on? http://senseis.xmp.net/?ShimadasKo? There are any number of what are termed http://senseis.xmp.net/?Superko that attempt to remedy situations like this and prevent non-terminating games (which are extremely uncommon but historically have happened, and usually get called as a draw/"no result"). The most straightforward one is to prevent repeated board configurations, this is called the positional superko rule. With positional superko (and with most superko rules), there are no legal non-finite games. A 3 move cycle not banned by the basic ko rule would be easy to construct on a much smaller board than 19x19, so if you are trying to steer them to the infinite situation I'd go with 9x9. However, if you have a positional superko rule (or the like), then the number they are trying to estimate will be finite, and so will be affected by board size. But even so I suspect the number is going to be very large even on a smaller board, and is going to involve lots of really complicated near-cycles that just slightly avoid the superko rule in use by one piece or so, and 9x9 will be big enough for this to be a really big number. http://senseis.xmp.net/?LongestPossibleGame along with a citation, but there seems to be a dispute about accuracy (I haven't tried to check it or follow up on the citation). There are at least examples of the kind of near-cycles I have in mind.

For your prime number question, it seems like you might want to show them trial division. This is a concrete way of understanding the concept. Let n=2^57,885,161 − 1 and systematically test whether each prime number less than n goes into n without a remainder. Count the number of prime numbers you have tested. From the way your question is written, it sounds like you already know that this is a bad idea, but, since mathematicians are never afraid to state the obvious: this is a bad idea. I mean, it might be a good idea for a smaller n, but not for this one. There's no way for your students to ever, in the amount of time remaining until the heat death of the universe, do enough trial divisions to count the number of primes smaller than this number. For this value of n, the prime number theorem is indeed the way to go, as people have mentioned upthread. http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument is a very interesting example of how mathematicians deal with infinite quantities/sets. It proves that there are different sizes of infinities: in particular, it shows that the infinite set of real numbers is strictly larger than the infinite set of natural numbers. It's somewhat of an advanced concept, but my high school math teacher taught us about this (also in an after-school advanced math program), and if you explain it well, it's totally understandable to students at that level. Another way to illustrate infinity and how arithmetic with infinite quantities differs from ordinary arithmetic is http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel. The idea there is that you have a hotel with (countably) infinitely many rooms, all of which are occupied, and you ask questions like: what do you do if another guest shows up? Well, just get the person in room 1 to move to room 2, the person in room 2 to move to room 3, and so on. So room 1 is empty, and the new guest can go to room 1. This shows that ∞+1=∞. Then what if (countably) infinitely many people show up? Well, ask the person in room 1 to move to room 2, the person in room 2 to move to room 4, ... , ask the person in room n to move to room 2n, etc. Then all of the odd-numbered rooms are free, so each of the new guests can be accommodated. This shows that ∞×2=∞. Then what if infinitely many buses, each containing infinitely many people, show up? And so on...

This may be of interest to you: "http://waitbutwhy.com/2014/11/1000000-grahams-number.html".

D) The prime number theorem states that the number of primes less than or equal to n is asymptotic to n/log(n). In other words, if p(n) is the number of primes less than or equal to n, then, as n goes to infinity, the ratio p(n)/[n/log(n)] goes to 1. You're interested, more or less, in p(2^a) for a certain a (in fact, you're interested in p(2^a-1), but that's the same as p(2^a) since 2^a is definitely not prime when a>1). The PNT tells you that p(2^a) is around 2^a/log(2^a), or 2^a/(a*log(2)). (Here "log" means "natural logarithm".) Now, the base-10 logarithm of 2 is like 0.3, and the natural logarithm of 2 is like 0.7, so, when a=57885161, you get 10^(0.3*57885161)/(0.7*57885161). So the right order of magnitude is like 10^17365540. In other words, it's dramatically bigger than anything else on your list. There is no way you're going to be able to actually count the primes less than 2^(58 million) using trial division. However, Beethoven's Sith's answer gives all kinds of fun questions you could ask: suppose every atom in the universe were actually a CPU core; how long would it take to count all the primes less than 2^57885161?, etc. There are some slightly more precise statements, though (I'm just looking at http://en.wikipedia.org/wiki/Prime_number_theorem#Statement since IANAANT). For example, for n in the range you're talking about, p(n) is between n/(log(n)+2) and n/(log(n)-4), but this extra accuracy doesn't really make the point any more strongly.

For your prime number question, it seems like you might want to show them http://en.wikipedia.org/wiki/Trial_division This is a concrete way of understanding the concept. Let n=2^57,885,161 − 1 and systematically test whether each prime number less than n goes into n without a remainder. Count the number of prime numbers you have tested. If you know any programming, you can write an algorithm to do this as is shown in the Wikipedia link. http://www.scientificamerican.com/article/infinity-logic-law/ is a recent SciAm story about how mathematicians think about infinity. And you could show them the documentaryhttp://topdocumentaryfilms.com/dangerous-knowledge/

So a cheap version of the prime number theorem mentioned above is that the ratio of pi(x) (the number of primes less than x) and the function x/ln(x) tends to 1 as x grows towards infinity. That means that although the difference between these two numbers can grow large, the relative error tends towards 0. So we can say that 2^57,885,161 − 1 / ln (2^57,885,161 − 1) is pretty close to the number of primes less than 2^57,885,161 − 1. So how can we get a sense of how big that number is? You can convert between the base of logs using this formula: log (base a) (x) = log (base b) (x) / log (base b) (a). Ln is log with base 'e', so let's convert to base 2 to make it easier to get the log of the power of 2: ln(2^57,885,161 − 1) = log2 (2^57,885,161 − 1) / log2(e) log2(2^57,885,161 − 1) will be very close to log2(2^57,885,161) which is just 57,885,161. log2(e) = 1.443. So the expression evaluates to around 40114456. That's around 2^25. Now divide the original number by 2^25 and we get 2^(57,885,161 - 25) = 2^57,885,136. How big is that in powers of ten? 10^3 is close to 2^10. 2^57,885,136 ~ (2^10)^5,788,514 ~ (10^3)^5,788,514 = 10^17,365,542. So, a lot of primes. (can someone check my work please!)

Fermi problems are really fun. A professor of mine in grad school used to give us these in our policy class. For instance, you can ask "How many pediatricians are there in Detroit?" It is possible, just by guesstimating several intermediary numbers (how many people in Detroit, what is the proportion of children to overall population, how many visits do children make on average, what is the maximum number of visits a single pediatrician can handle, etc.) to come up with a number that is within an order of magnitude of the correct answer. There is not necessarily a correct way to get to this answer, which is what makes it so fun (and the steps I showed above aren't necessarily the right ones; I can't remember how we solved it exactly). But it's surprisingly possible--as well as great fun--to get from here to there... it sounds like you've got smart kids and would like to challenge them, so they might love the challenge of these sorts of problems...

2^57,885,161 − 1 / ln (2^57,885,161 − 1) A great example of the type of question you are looking for, is exactly: How would you evaluate a formula like this, involving very large numbers? It ain't easy! Other equations evaluated at very large numbers could be quite a lot easier, say x^3 -x^2 + 10x - 1000 evaluated at x=10^100. (At large numbers the x^3 dominates so you can just ignore the other terms for practical purposes.) Others can be fiendishly difficult: What is tan(x) for x=10^100? That might be very difficult to calculate to even the slightest degree of precision. And that was the easy one--now how are you going to calculate tan(2^57,885,161) or (even better) tan(2^57,885,161^57,885,161). BTW at least some of the solutions to the problems outlined above are in the reach of students at this level, at least conceptually. All you need to figure tan(x) is to know the value of x mod pi. So to figure out 10^100 mod pi, you need to know the value of pi to some hundred-odd digits, then be able to do arithmetic operations on numbers with that many digits. So, doable. 2^57,885,161 mod pi is a lot harder because you're going to need to know pi to some 10s of millions of digits and be able to do arithmetic on numbers with that many digits. 2^57,885,161^57,885,161 mod pi is perhaps not something that is knowable, given the age of the universe, fundamental limitations imposed by entropy, etc. Unless someone comes up with a clever solution that bypasses all that, of course! This little exercise is nice because it ties in to the question of in what possible way calculating pi to a million or hundred million digits could be useful. One simple answer is, you can't figure out a simple question like what is tan(x) unless you know pi to the same number of digits x has, plus some.

BTW what made me think of tan(10^100) is that it is the punchline of a famous story told by http://lesswrong.com/lw/cvf/where_fermi_fails_what_is_hard_to_estimate/ about his prowess at solving Fermi-type problems. It's a http://lesswrong.com/lw/cvf/where_fermi_fails_what_is_hard_to_estimate/ and might give you some ideas to think about.