From: Ching Kong Newsgroups: sci.math.num-analysis Subject: Re: Here's a tough puzzle... Date: Thu, 19 Sep 1996 00:29:58 -0500 Since the bases and operators are finite, this puzzle can be solved by a brute force method. According to the question, the basis is 4. With unitary operations, i.e., n!, and SQR(n), we now have v1= < 4, 24, 2 > as our basis. By using binary operations, i.e., +, -, *, /,and ^ we can generate the following matrix: + 4 24 2 4 8 28 6 24 +/-20 48 26 2 +/- 2 +/-22 4 0 Similarly, * 4 24 2 4 16 96 8 24 6 576 48 2 2 12 4 1 And so on... So by using two 4's, we have v2=<-22, -20, -2, 0, 1/12, 1/6, 1/2, 1 2 4 6 8 12 16 20 22 26 24 26 28 (44) 48 96 256, ... 24^24> as our basis. (After sorting, and ignoring SQR(v2) and (v2)! for simplicity.) By using v1 and v2, we can proceed to generate v3 with the same method. v3 should consist all the possible numbers that can generate from three 4's with the operators mentioned. We then can generate all the possible number that can form from four 4's and the operators by v4=SORT(operator(v3,v1)/\operator(v2,v2)). (/\=logical AND) (ignoring (v4)!, SQR(v4), ((v4)!)!, SQR(SQR(v4)), ... etc) To save time, I found SORT(+(v2,v2)/\-(v2,v2)/\*(v2,v2)) will generate all the number between 0-> 100 except: 11 15 19 23 31 33 35 37 39 41 43 51 53 55 57 58 59 61 62 63 65 67 69 71 73 75 77 78 79 81 82 83 84 85 86 87 89 90 91 93 95 and 99. (That's more than 50 generated.) 62 and 90 belong to operator(v3,v1): 4*4*4-2=62 and (4!/4)/4/SQR(4)=90 and other numbers in the 'miss-list' is also belong to operator(v3,v1). Those numbers that do not appear in v4 and unitary operations on v4 can NEVER be generate by four 4's and the operators. Mr. Zabel don't think anyone is awesome. ;) On 18 Sep 1996, Zabel wrote: > Try this out: > > Use four 4's and add, subtract, multiply, divide and square root them to > try to come up with every number from 1 to 100. You must use all four > 4's, and no more than four. You can not square any of the fours to make > it 16, but you can raise a four to the fourth power. > > You can also use factorials. Like (4! * 4 + 4) / 4 > 4! is equal to 4 * 3 * 2 * 1 which equals 24. > > and you can use 4 to the 4th power. <-- This counts as using 2 fours. > > You put the square root sign over any part of the equation. It is > hard to illustrate on the computer, but I hope you can see what I'm doing > here: > (square root of (4) * 4!) / (square root of (2)) + 4! = 72 > > > This is how you get the number 1: > (4 + 4 - 4) / 4 = 1. > Now try to figure out how to get the number 2. Then 3. Up to 100. > It is possible. Some are very easy, some are medium, and some are very > difficult to figure out. > ... > > You can also group the fours into the numbers 44, 444 and 4444. But if you > use 4444, then you've used up all your fours. > > If you can get 50 or more, E-Mail them to me. > If you get all 100, you are awesome. > > Ian Zabel > zabel@planet.net > http://planet.net/pzabel > > > > >