From: edew@netcom.com Newsgroups: sci.math Subject: Re: Help with four 4's Date: 18 Jan 1996 22:19:02 GMT In article jbrown@epix.net (J Brown) writes: =In article <4d8iod$ltj@acme.freenet.columbus.oh.us>, =dsayre@freenet.columbus.oh.us (Denise Sayre) wrote: = => I was given this problem for extra credit in my math class: => => Number your paper from 0 to 100. Now using only four 4s for each => number, use the functions (+,-,*,/,sqaure root (no other roots except => the quadruiple root, but that counts as one of youre fours), factorial, => exp (again only to the fourth power, and again counts as a four),decimals, => and fractions) to create each number on your list. You can only use four => 4s, no more or no less. 44 counts as two 4s, 4.44 counts as three, etc. = => ANY help on odds or answers would be appreciated. => => Thank you very much. => -- => Andrea Miller dsayre@freenet.columbus.oh.us = =This is a very much used problem in mathematics classes everywhere. =My eighth grade students worked on it earlier in the year, we did one to ten. =Did you try using 4 divided by 4? that will give you a 1. =1 = 4/4*4/4 =3 = sqrt4 + sqrt4 - (4/4) =5 = sqrt 4 + sqr4 + (4/4) = =That should help you get started.... write again if you need more help. = =My class is currently working on a similar problem using the digits in the =current year 1996 instead of four fours. We do not allow the use of =double digit numbers such as 19 or 99 or 96, but all the other rules are =the same. We know (from last year) that it is possible to express the =numbers 1 through 100 using the digits in the year 1995, but we do not yet =have all the numbers for 1996. You might suggest this problem to your =teacher. = =Judy While as an undergraduate at UCSB (1984), the women's volleyball team held a fund-raising drive by offering a $10,000 prize to anyone who can find out the three numbers between 1 and 1000 that cannot be expressed with four 4's. I didn't pay, but I tried for a while. It's amazing how many numbers can be written using four 4's. I recall the rules allow sqrt, 4... (or .\overline{4}), factorial, and all the regular things. Now, I can't even do up to 30. It was fun. I asked one of my math profs and he was really excited about the $10,000 prize. Of course, I don't know how anyone can prove a number can not be expressed as some algebraic expression with four 4s. EDEW ============================================================================== From: David Shield Newsgroups: sci.math Subject: Re: Help with four 4's Date: 15 Jan 1996 07:05:34 GMT dsayre@freenet.columbus.oh.us (Denise Sayre) wrote: > > I was given this problem for extra credit in my math class: > > Number your paper from 0 to 100. Now using only four 4s for each >number, use the functions (+,-,*,/,square root (no other roots except >the quadruple root, but that counts as one of your fours), factorial, >exp (again only to the fourth power, and again counts as a four),decimals, >and fractions) to create each number on your list. You can only use four >4s, no more or no less. 44 counts as two 4s, 4.44 counts as three, etc. > > Evens are >possible, but the odds are...overwhelming. (snip) > Thank you very much. >-- >Andrea Miller dsayre@freenet.columbus.oh.us > I have, in the past, spent some time on this puzzle, but have come nowhere near getting all the numbers up to 100. I have got up to 30, but then have gaps at 31, 33, 37, 39, and most of the rest of the way! And the only way I have found to represent 40 would not be allowed under your rules; it used the function [x] = integer part of x (e.g., [3.75] = 3, [3] = 3). That was: 40 = [(4^4)/(4!)]*4 No, if there is any easy pattern, I didn't find it! You speak of the even numbers as being more do-able; did you find an expression for 40? Admittedly, I did not use square roots, which could have been useful: 46 = 44 + 4 - sqrt(4), was not on my list before. > My teacher says its possible but after we did >it, and turned it in he said he didnt know the answers. I would very much >like to show him. Yes, I can understand that! I wonder what was his basis for saying that all numbers to 100 are possible? He might be right, of course, but my answers are pretty sparse from 50 on. I felt quite proud of 90 = ((4!/4)!) / (4+4) (of course 45 = ((4!/4)!) / (4*4), but 45 = 44 + 4/4 is so much simpler.) ..and a nice finish is: 100 = (4!+(4/4))*4 I'm not very familiar with the internet. Is an answer to a popular puzzle like this already accessible somewhere? David (d.shield@bendigo.latrobe.edu.au) ============================================================================== From: b.scott@bscott.async.csuohio.edu (Brian M. Scott) Newsgroups: sci.math Subject: Re: Help with four 4's Date: 16 Jan 1996 16:02:23 GMT In article <4de19d$gjm@math.mps.ohio-state.edu>, stadler@math.ohio-state.edu (Jonathan Stadler) says: >Yes, it is possible to get all 100 numbers. However, in order >to do so, I know that I had to use the greatest integer function. I think that one does need more than the constructions mentioned in the original post, but there's a more elementary notation than the greatest integer function that will do the trick: the old overbar notation for repeating decimals, e.g., _ .4 = .444444... = 4/9. Writing this as {4} for convenience, one has 4/{4} = 9 and other useful representations. (Some numbers are still tough, though; it took a while to find (4! + 4! + sqrt({4}))/sqrt({4}) = 73.) Brian M. Scott Cleveland State Univ. scott@math.csuohio.edu ============================================================================== From: Gunnar R Grape Newsgroups: sci.math Subject: Re: Help with four 4's Date: Wed, 17 Jan 1996 22:48:34 -0800 Hi folks, hi Brian and Colin ... Colin Percival wrote (among other things): _ >33 = 4! + (4/.4)^(sqrt(4)) _ >(.4 means 0.44444...) >I've managed to get all the numbers from 1 to 100 with factorial,square >root, +, -, *, /, decimals and concat. Brian Scott uses the same notation (73 was elegant, by the way). We have to get the rules straight here! Are zeroes allowed (eg 0.4)? Is concat allowed? In that case I have a feeling it would be a piece of cake! Concat was not mentioned originally. In the notation _ .4=0.444 ... you are actually using a zero and an untold nr of fours. _ By the way, since .4 = 4/9, the expression for 33 was slightly wrong. _ It could read 33=4!+sqrt(4)*sqrt(4)/.4 (assuming it is legal ...) Not using any fancy stuff (only + - * / ^ factorial sqrt), I've gotten everything except 26 odd nrs from 33 on up, not having spent that much time on it. This kind of exercise is fun, but a strict set of rules is necessary! Who calls this game? Keep truckin', gang! Gunnar ============================================================================== From: Leo.Willemsen@nlrtdfsc.origin.nl (Leo Willemsen) Newsgroups: sci.math Subject: Re: Help with four 4's Date: 17 Jan 1996 08:11:29 GMT In article <4dgu15$lm4@morgoth.sfu.ca>, cperciva@sfu.ca says... > _ >33 = 4! + (4/.4)^(sqrt(4)) > _ >(.4 means 0.44444...) > >Colin Percival your .444 looks a bit like cheating to me :-). I got stuck on 31 for which you found a neat solution. I found a 'cheat': 31=4!+4+4-isqrt(4) where isqrt means: take the square root infinitely often as lim (n->inf) nth root of any number = 1. But I guess, that should be disallowed as well. -- 1 Leo Willemsen ============================================================================== Newsgroups: sci.math From: phil@eurocontrol.fr (Phil Gibbs) Subject: Re: Help with four 4's Date: Tue, 16 Jan 1996 14:03:01 GMT > I got to 32 (no sqrt), but I got stuck on 33. > By the way, 31 = 4!+(4!+4)/4. 40 = 4!+4!-4-4. > I'm sure lots more are possible, but I doubt that all are. Unless I've > misunderstood the rules. Of course one could write a little computer > program to look into the set generated by all well-formed expressions, > and be done with the problem ... How far you can get depends on what operations you allow. A useful notation is one which allows you to write recurring decimals e.g. .(4) = 4/9 If you allow this notation and square roots etc, it is possible to get up to 112. I dont know what the result is if you dont allow square roots or recurring decimals but here are some possibilties for 33 with sqrt. 33 = 4! + 4 + sqrt(4)/.4 or 33 = (sqrt(sqrt((4^(4!)))) + sqrt(4))/sqrt(4) There are more possibilities than you think, you just have to be inventive, that is what the makes the game so interesting. If you allow square brackets for "integer part of" it becomes too easy and less interesting. for solutions to the puzzle with the digits 1,2,3,4 look at http://ourworld.compuserve.com/homepages/Phil_Gibbs/1234.htm ============================================================================== From: shayd@post.tau.ac.il (David Shay) Newsgroups: sci.math Subject: Re: Help with four 4's Date: Tue, 16 Jan 1996 19:30:56 GMT dsayre@freenet.columbus.oh.us (Denise Sayre) wrote: > I was given this problem for extra credit in my math class: > Number your paper from 0 to 100. Now using only four 4s for each >number, use the functions (+,-,*,/,sqaure root (no other roots except >the quadruiple root, but that counts as one of youre fours), factorial, >exp (again only to the fourth power, and again counts as a four),decimals, >and fractions) to create each number on your list. You can only use four >4s, no more or no less. 44 counts as two 4s, 4.44 counts as three, etc. > I did not fair well on his extra credit assingment. Evens are >possible, but the odds are...overwhelming. Is their a pattern to the odd >numbers. If so what is it. Is the a way to make a 4 into an odd number >using only one function. Have any of you heard of this problem before. >Are the answers anywere on the net? Does anyone have them? If so, could >you please mail them to me. My teacher says its possible but after we did >it, and turned it in he said he didnt know the answers. I would very much >like to show him. ANY help on odds or answers would be appreciated. > Thank you very much. >-- >Andrea Miller dsayre@freenet.columbus.oh.us The book "Mathematical Bafflers", edited by Angela Dunn, Dover Publications, New York, 1980, contains a full solution to your problem (page 4). David Shay ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Help with four 4's Date: 19 Jan 1996 20:26:24 GMT In article <4d8iod$ltj@acme.freenet.columbus.oh.us>, Denise Sayre wrote: > > I was given this problem for extra credit in my math class: > > Number your paper from 0 to 100. Now using only four 4s for each >number, use the functions (+,-,*,/,sqaure root (no other roots except >the quadruiple root, but that counts as one of youre fours), factorial, >exp (again only to the fourth power, and again counts as a four),decimals, >and fractions) to create each number on your list. You can only use four >4s, no more or no less. 44 counts as two 4s, 4.44 counts as three, etc. Ask to include logarithms (any base) instead of, say, factorials. For any n, use n sqrt's in the following formula: log[log(4)/log(sqrt(sqrt(...(sqrt( [sqrt(4)*sqrt(4)] ))...)))]/log(sqrt(4)) This then comes out to log[log(4)/log(4^((1/2)^n))]/ log(2) =log( 2^n )/log(2) =n There's no need to stop at 100, then! dave ============================================================================== Newsgroups: rec.puzzles,sci.math From: reeds@research.att.com (Jim Reeds <4027-29348> 0112110) Subject: Re: making numbers with just one 4 Date: Fri, 19 Jan 1996 23:17:01 GMT Note that sqrt(sqrt(...sqrt(4)...) = 4 ^ ( 2 ^ -k ), where x^y means x raisd to the y power, where there are k nested sqrt() functions. Hence ln(ln(sqrt(...sqrt(4) ...) = -k ln 2 + ln ln 4. Denote this quantity b, which uses just one 4 to write, by a(k). Then any positive rational number u/v can be written with exactly four 4s because u/v = ((a(1)-a(u+1))/((a(1)-a(v+1)). -- Jim Reeds, AT&T Bell Labs, Room 2C-357 600 Mountain Avenue, Murray Hill, NJ 07974, USA email: reeds@research.att.com, phone: +1 908 582 7066, fax: +1 908 582 2379 home page: http://netlib.att.com/math/people/reeds/index.html ============================================================================== Subject: Returned mail: Host unknown Date: Wed, 24 Jan 96 10:09:01 CST From: rusin (Dave Rusin) To: e0felr94@tuzo.erin Subject: Re: 1996 Puzzle Newsgroups: sci.stat.math,sci.math In article you write: > > I have recently been working on a puzzle that requires the generation >of all numbers between 1 and 100 using the digits 1, 9, 9, 6. I have been >able to get them all with the exeption of 68. Is there anyone that can >obtain this number? Any mathematical functions are allowed, as long as ^^^^^^^^^^^^^^^^^^^^^^^^^^ Well, among the many functions available to mathematicians are the constant functions! Try the function f which is defined by f(x)=68 for all numbers x Too boring? How about the Successor function: S(x)=x+1 for all numbers x. Then 68=S(S(S(...S(1+9+9+6) ...))) (insert as many S's as needed). Still too ridiculous? Well here's my recently-posted response to a similar problem: [my 19 Jan post, above, was quoted; deleted 2/29/96 --djr] The point of this message is that the problem is too poorly defined to admit any reasonable solution. It is a valid question, for example, to ask if you can combine the four numbers (not "digit") 1, 9, 9, and 6 using + - * / only so as to make 68. This is a finite problem which a computer can check; the answer is no. dave