Date: 2003-Oct-29 14:27:50 <preambule> Sorry for my poor English, it is not my native language. It is not even my second natural language. :) </preambule> Here is an easy puzzle (easy if you know the solution, of course). Using any mathematical operations and only three numbers "2" express any natural number. For example, using only two "2" you may express 4 as: 4 = 2^2 or 4 = 2*2 or 4 = 2+2 Returning to our task, a naive approach shows: 0 = 2 * (2 - 2) 1 = 2 ^ (2 - 2) 2 = 2 + 2 - 2 3 = 2 + 2 / 2 4 = sqrt(2 + 2) + 2 5 = ? 6 = 2 + 2 + 2 7 = ? How would you express 2003? If you know the answer, please don't post it for several days.

Date: 2003-Oct-29 15:07:09 I'd say there is probably more than one solution to this problem, since you could simply write: 2^(2-2) + 2^(2-2) + 2^(2-2) + . . . {repeat 2003 times} Or alternatively 2 + 2 + 2 + ... {repeat 1001 times} + 2^(2-2) Or 2*2 + 2*2 + 2*2 + ... {repeat 500 times} + 2 + 2^(2-2) Etc. Are there other constraints?

Date: 2003-Oct-29 15:11:37 Sorry, I just reread the problem and say that you are limited to only 3 '2's, ignore my previous post.

Date: 2003-Oct-29 17:05:59 I don't know if this solution counts, since it kind of feels like cheating... First define the function f(x) := x+(2/2), which uses two 2's. From this, any natural number >=2 can be expressed using the third 2, in the form f(f(...f(f(2))...)) Is this the solution you were looking for? If not, then exactly which mathematical operations are we allowed to use? Mike

Date: 2003-Oct-29 17:26:54 Mike: your solution is nice, but it uses some unknown mathematical function f(), my solution uses only known mathematical operations. :)

Date: 2003-Oct-29 17:30:20 Let's call the original problem "2.2.2.c". If "2.2.2.c" is too hard for you, solve these easier problems. Problem "2.2.2.a" ================= How many positive integers (including 0) can you express using some mathematical operations and three numbers "2" only? I already expressed 6 numbers, here are two additional expressions: 8 = 2 * 2 * 2 = several more ways 16 = 2 ^ (2 + 2) = (2 + 2) ^ 2 = 2 ^ 2 ^ 2 = several more ways Are there other numbers except for these seven (the expression for 4 uses sqrt so it does not count)? You may use only operators "+", "-", "/", "*", "^" and grouping. Problem "2.2.2.b" ================= The same like the previous, but you may join digits, i.e. 22. You may also use any reasonable mathematical functions, like sin() or sqrt(), but you can't use such unreasonable functions like floor() or rand(). You are also not allowed to use factorial for simlicity. Post a list of additional numbers that you were able to express this way. P.S. So, what is the difference between "2.2.2.b" and "2.2.2.c"? None, except that "2.2.2.c" includes a solution for "2003" and you are not expected to post it until November 2. :)

Date: 2003-Oct-29 21:10:21 Well, I'm *SURE* this is not the intended solution, but an answer is simply: 222 So long as you are using base [sqrt(4003)-1]/2. :-) (The positive solution to the equation 2x^2 + 2x + 2 = 2003)

Date: 2003-Oct-29 22:53:52 I *SURE* hope that's not the answer, it was one of the first ideas I had, but immediately dismissed because of "being too disappointing/unrewarding". In the same style -> 22E-2 Peter

Date: 2003-Oct-29 23:04:02 sqrt(x) = x^(1/2) = x^(2/(2*2)), which eats up all of your twos. ;-) There must be some trick to this. Can you give an example of how to do it for some number > 1000 (but not 4194304, heh), or would that give it away?

Date: 2003-Oct-29 23:26:50 Ok, I may give one hint (I will add one small hint every day). :) The solution for "2.2.2.c" does not use joining like 22, i.e. separate numbers 2 only, so I guess your tricks with a different base can't work. But what about "easier" problems? We may skip obvious integers 11, 20, 24, 44, 222, 484 and 4194304. Can you produce 5 or 7, for example? ;-)

Date: 2003-Oct-30 09:08:35 > The solution for "2.2.2.c" does not use joining like 22, i.e. separate > numbers 2 only, so I guess your tricks with a different base can't work. I can verify that joining of 2s is in fact not used in the solution. ;-) > But what about "easier" problems? > We may skip obvious integers 11, 20, 24, 44, 222, 484 and 4194304. > Can you produce 5 or 7, for example? ;-) Now I can! Neat puzzle. At first I thought it was impossible, and that a "solution" would only exist due to a clever play on words, or some advantageous bending of the rules. It turns out that there really is a mathematical solution, and several of you are more or less on the right track to finding it. That's all I'll say for now, lest I give too much away... ;-)

Date: 2003-Oct-30 09:48:54 Does it include - recursion - integration - summation/product - stuff with variables x or n ?

Date: 2003-Oct-30 11:07:27 jason: I am glad someone solved it. After all this is the MATH room. To be honest, I don't know whether I would solve it myself, but I try my best to give hints. You see, my hint about producing 5 or 7 helped. :) You already mentioned my next daily hint. The incorrect solutions suggested by some people are not quite in the right direction, but they are also not in the opposite direction, so investigating them may help. plewylli: I have a counter question. Does all of these help you? If yes, use it. :) However, the intended solution does not use anything like this.

Date: 2003-Oct-30 11:25:08 Yesterday Peter sent me this totally incorrect, but still nice solution. :) >> floor(sqrt(exp(ceil(sqrt(exp(2)))))) ans = 4 >> floor(exp(exp(2)))+ceil(ceil(exp(4))*floor(exp(2))) ans = 2003 I post it here so others don't try to do it this way.

Date: 2003-Oct-30 16:52:16 What operations are allowed exactly? I would have thought that exp (or e^x) would be off limit, since e is a number as well. ln x I can see (log base e), as well as log x (log base 10), but lg x is also commonly used (log base 2). If the last is allowed, 5 is easy: lg(2) + 2 + 2 So again...what is 'legal' in the problem?

Date: 2003-Oct-30 17:00:30 This one is like too easy...

Date: 2003-Oct-30 17:00:16 Following up on what I just said, (lg(2)+2)! - lg(2) = 5 (lg(2)+2)! + lg(2) = 7

Date: 2003-Oct-30 18:51:46 Believe it or not, I think I've found a way to express 2003 using ONLY TWO 2's, without using any dubious functions like floor, ceiling, or round. (I can also do it with three 2's, but the two-2 solution is nicer.) So here's a question for the other people who've solved it with three 2's: Does your method work with only two 2's? In any case, I won't post my solution yet. Mike

Date: 2003-Oct-30 18:55:15 > So here's a question for the other people who've solved it with three 2's: > Does your method work with only two 2's? Technically, yes... ;)

Date: 2003-Oct-30 21:52:58 tomc: Finding the mathematical operations to be used is a part of the task. In my next hint I planned to say how many different operations are used in my solution, and in the hint after that mention one or two of them. But, ok, I will give you a better hint. :) If we suppose your equations for 5 and 7 are true (something that is quastionable), then it is possible to express 2003 using ONLY ONE 2.

Date: 2003-Oct-30 22:17:01 > If we suppose your equations for 5 and 7 are true (something that is > quastionable), then it is possible to express 2003 using ONLY ONE 2. It occurred to me after I made my previous post that my solution method does indeed make it possible to express 2003 using *ONLY ONE 2*. (Had the problem been stated this way initially, I would have thought it truly impossible...)

Date: 2003-Oct-31 03:34:01 The daily hint. If you assume that the following expression 6 = 2 + 2 + 2 uses only one different mathematical operation, then the intended solution for this puzzle uses only 3 different operations. To answer Tom's questions, if you want to use logarithms in your solution, then you may only use standard ln and log (natural and decimal logarithms) implicitly. If you need another base you should specify it explicitly. Otherwise such solution would be ambiguous, I guess.

Date: 2003-Oct-31 13:36:24 I wouldn't expect an arbitrary base logarithm, no. But 'lg' is a standard notation for log base 2. You see it commonly used in discussing big O notation analysis for algorithms, such as mergesort being N lg N, etc. That's the reason I suggested it. Of course, even accepting that as a possible answer, that doesn't really help with the 2003 problem, only the 5 and 7 problem. (Putting 'N lg N' into google gives several eexamples, such as: http://www.cse.ucsc.edu/classes/cmps012b/Spring97/Lecture14/sld005.htm) So the problem is to find an acceptedly common function (nothing custom) which yields 2003 as an answer - but that means that 'acceptedly common' is somewhat arbitrary, since we've seen 'ceil' and 'floor' both used successfully. Some operations to choose: Clearly accepted: + - * / ^ ! sqrt abs Possibly accepted: log ln lg trig functions (and anti-functions) hyperbolic functions (and anti-functions) exp (e^x) modulus Stretching: floor ceil (unless it was the 'exp' that was the problem) fibonocci(n) (nth number in the Fibonocci sequence) triangle(n) (nth triangle number, i.e., n(n+1)/2) ------------------------------------------------------------ As an interesting side note, testing nearly random things, I came across this: pi/2 - atan(2003) ~= 1 / 2003

Date: 2003-Oct-31 18:08:15 Tom, you may want to consider the trigonometric functions as "clearly accepted", as suggested by one of migo's earlier posts: > You may also use any reasonable mathematical functions, like sin() or > sqrt(), but you can't use such unreasonable functions like floor() or > rand(). You are also not allowed to use factorial for simlicity. Another hint I picked up from this earlier post of his is that the factorial function isn't needed, at least not for the intended solution. Also, to expand on migo's hint of the day that the intended solution uses 3 distinct operations, I'll add that there is also a solution which uses only 2 distinct operations.

Date: 2003-Nov-01 04:07:49 This is the last hint. Tomorrow I will post my solution. The three mentioned operations are log2 (log base 2), sqrt and "-". If you insist on using lg to mean log2, then the puzzle is to express any number using only one "2", otherwise it is still three "2". I think no more hints are possible, you _must_ know the answer now. :) I didn't have a free minute today (just released new versions of two free software projects) to think about the Mike's solution with only two operations.

Date: 2003-Nov-01 07:02:49 Michael wrote: > Also, to expand on migo's hint of the day that the intended solution uses 3 > distinct operations, I'll add that there is also a solution which uses only > 2 distinct operations. then Mikhael wrote: > The three mentioned operations are log2 (log base 2), sqrt and "-". > > If you insist on using lg to mean log2, then the puzzle is to express > any number using only one "2", otherwise it is still three "2". Ahh, nice... I think I just discovered Mikhael's solution, and I believe it's probably the same as Michael's 2-distinct-operation solution, except Mikhael is also counting the '-' as an operation. It is similar to, but more elegant than, the first solution I found, which uses sqrt(), and logarithms in two different bases. I suspect this is the same method used in Michael's 3-distinct-operation solution.

Date: 2003-Nov-01 21:13:07 Whoah, migo's solution is *really* different than mine---I'm not using log2, sqrt, OR "-"! Instead, I'm making use of 2 trig functions. Maybe I'll try doing it with the log/sqrt method if I have time before tomorrow. -- Michael

Date: 2003-Nov-02 01:25:07 It's getting interesting. I think I'll wait a bit with posting my solution. I doubted whether to disallow trigonometry functions from the start. Because it is always ambiguos, we should know whether to work in radians or grads or whatever. I thought, trigonometry functions can't help anyway, so it is obvious they can't be used in the solution. :) Michael, if you want others to find your solution, formulate it properly. Is it to express any integer number using only one number 2 and only two distinct trigonometry functions working in radians? I can't see how more than one "2" may be used, since all trigonometry functions I know accept only one argument. Does posting the function names reveal the solution? :)

Date: 2003-Nov-02 05:15:36 Okay, I've figured out a nifty solution using "-", sqrt, and log2. I'll let Mikhael post this solution when he's ready. If anyone wants to try figuring out my solution, the problem is this: Express any integer >= 2 using only one 2, and up to two distinct operations. (As I already hinted, I happened to use trigonometric functions, but there may be non-trigonemetric approaches as well. The trigonometric functions I would consider valid are sin, cos, tan, csc, sec, cot, and their inverses (asin, acos, atan, acsc, asec, and acot). You can work in whatever angle system you want (radians, degrees, or gradients). I know of 2 solutions.

Date: 2003-Nov-02 09:21:32 The original solution follows. Very elegant. The tall thing is sqrt(). To express 2003, use 2003 square roots. _____ /_____ //_____ ///_____ ////_____ /////_____ //////_____ 7 = - log log VVVVVVV 2 2 2 You may verify this by typing "perl" in your shell and pasting this code: sub log2 { log($_[0]) / log(2) } # perl does not have log_2, only log_e print -log2(log2(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2)))))))));

Date: 2003-Nov-02 19:23:05 Similar to Mikhael's original solution, given a natural number n, use n iterations of sqrt(): n = log log 2 2 sqrt(sqrt(...sqrt(2)...))

Date: 2003-Nov-03 16:56:16 This problem was beautiful :) . Thanks.

Date: 2003-Nov-03 20:51:10 I tried to post before, but I guess the system went down when I was trying. The solution is quite elegant, I like it. There's one remaining question in my mind: how do you solve the problem using the trig functions, as mentioned before? I'd like to see that solution as well.