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Archive-name: puzzles/archive/pickover/part2
Last-modified: 17 Aug 1993 Version: 4 See reader questions & answers on this topic! - Help others by sharing your knowledge ==> pickover/pickover.07.p <== Title: Cliff Puzzle 7: 3x3 Recursion From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider the 3x3 array below. All nine digits are used exactly once. 1 9 2 3 8 4 5 7 6 Notice that "384" is twice the number in the first row, and that "576" is three times the number in the first row. Questions: 1. Are there other ways of arranging the number to produce the same result using each digit only once and the same rules? Remember, the second row must be twice the first. The third row must be 3 times the first row. 2. Start with the number in the last row (e.g "576" or any other solution you may find) and continue to form another 3x3 matrix using the same rules with the new starting number. In other words, the number in the second row must be twice the first. The third row must be three times the first. (For this problem you may truncate any digits in the beginning. For example, 1384 would become 384.) Keep going. How many matrices can you create before it is impossible to continue. Again, each digit must be used only once in each matrix. ==> pickover/pickover.07.s <== ------------------------- > Title: Cliff Puzzle 7: 3x3 Recursion > Consider the 3x3 array below. All nine digits are used exactly once. > 1 9 2 > 3 8 4 > 5 7 6 > Questions: > 1. Are there other ways of arranging the numbers to produce the same > result using each digit only once and the same rules? YES . 2 1 9 2 7 3 3 2 7 4 3 8 5 4 6 6 5 4 6 5 7 8 1 9 9 8 1 > same rules with the new starting number. In other words, > the last number becomes the first, and the number in > the new second row must be twice the first. The third row must be three > times the first. (For this problem you may truncate any digits in the > beginning. For example, 1384 would become 384.) NONE. There is no solution to the continuation problem. Bye. Amit Agarwal > to continue? Again, each digit must be used only once > in each matrix. > > ------------------------- By exhaustive search I found that there are only four such arrays. Here they are: 1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1 Since these are the only four it is clear from inspection that none of the last numbers ever begin another array with the desired properties. Bob Murphy (rmurphy@aludra.usc.edu) ------------------------- Well, I think I have an answer to both parts. I did what should have been a complete analysis of all possible column combinations, but it is certainly possible that I missed a point somewhere. If you don't get any answers contradicting this one, I'd be happy to send you my analysis. Anyway - for part 1, I found the following three matrices (in additionn to the one you gave): 2 1 9 2 7 3 3 2 7 4 3 8 5 4 6 6 5 4 6 5 7 8 1 9 9 8 1 Note that the first one of these can be generated from yours by moving the third column to the first position, and the third one can be generated from the second similarly. In both cases, one column does not receive or produce any carryovers, so it can be placed at either end. For part 2, there is obviously no solution, assuming that these are indeed the only four matrices satisfying the requirements. In my analysis, I included columns with carryovers in all positions, so if there were any matrices that would satisfy the relaxed condition of part 2 I should have found them. Dan Blum Institute for the Learning Sciences Northwestern U. blum@ils.nwu.edu ------------------------- Cliff, In article <1blrk9INN10s@aludra.usc.edu> (Bob Murphy) writes: >By exhaustive search I found that there are only 4 starting numbers >which produce a 3x3 array with the desired property. Here they are: > > 1 9 2 2 1 9 2 7 3 3 2 7 > 3 8 4 4 3 8 5 4 6 6 5 4 > 5 7 6 6 5 7 8 1 9 9 8 1 For each of these solutions I happened to notice that the sum of each row is a constant: sum(row1) = 12 sum(row2) = 15 sum(row3) = 18 (necessary but not sufficient condition) And the sums all differ by the same constant (3). I wonder if this property may somehow be generalized to matrices of higher degree? Regards, -- Greg Schmidt schmidtg@iccgcc.decnet.ab.com ------------------------- > If you respond to this puzzle, if possible please send me your name, address, > affiliation, and e-mail address so I can credit you in the future if needed. > If you like, tell me a little bit about yourself so I can cite you > appropriately if you provide unique information. PLEASE ALSO directly mail > me a copy of your response in addition to any responding you do in the > newsgroup. I will assume it is OK to describe your answer in any article or > publication I may write in the future, with attribution to you, unless you > state otherwise. > Thanks, Cliff Pickover > > Consider the 3x3 array below. All nine digits are used exactly once. > > 1 9 2 > 3 8 4 > 5 7 6 > > Notice that "384" is twice the number in the first row, and that > "576" is three times the number in the first row. > > Questions: > 1. Are there other ways of arranging the numbers to produce the same > result using each digit only once and the same rules? > Remember, the second row must be twice the first. The third row > must be 3 times the first row. > > 2. Start with the number in the last row (e.g "576" or any other > solution you may find) and continue to form another 3x3 matrix using the > same rules with the new starting number. In other words, > the last number becomes the first, and the number in > the new second row must be twice the first. The third row must be three > times the first. (For this problem you may truncate any digits in the > beginning. For example, 1384 would become 384.) > > Keep going. How many matrices can you create before it is impossible > to continue? Again, each digit must be used only once > in each matrix. Well, this is probably not news to you by now, but I only get four solutions to the original problem: 1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1 If we relax the rules slightly and allow zeroes, and just specify that the nine numbers only have to be different, then we get two more solutions: 0 7 8 2 6 7 1 5 6 5 3 4 2 3 4 8 0 1 both of which use the digits 0-8, which may be of interest. The second problem (in either form) has only the above solutions, with only one matrix in each solution. If we switch to base 9 (where we must use a zero), there is no solution to the first, and only one solution to the second (with only one matrix): 4 8 1 0 7 2 5 6 3 In fact, I considered 3 versions of problem 2. In all cases zeroes were allowed, but the 9 numbers had to be different. For each of them the first 3x3 matrix has to meet the original specifications; where they differ is in how the succeeding matrices are constructed. In the ensuing discussion, the original number is called 'n'. So in the example given with the problem, n is 192. Version A: The second matrix consists of rows with n*2, n*3 and n*4 in them. (The last three digits of those, anyway.) The next would have n*3, n*4, and n*5, then n*4, n*5, n*6, etc. Version B: The second matrix consists of n*3, n*6, n*9. (This is essentially the second problem as given.) Version C: The second matrix consists of n*4, n*5, n*6. The next would have n*7, n*8, n*9 etc. Results for various bases: Base 9: A, B, C: 4 8 1 0 7 2 5 6 3 Base 10: A, B, C: 0 7 8 1 9 2 2 1 9 2 6 7 2 7 3 3 2 7 1 5 6 3 8 4 4 3 8 5 3 4 5 4 6 6 5 4 2 3 4 5 7 6 6 5 7 8 0 1 8 1 9 9 8 1 In addition, with version C, we get a second matrix for 219. 2 1 9 8 7 6 4 3 8 ==> 0 9 5 6 5 7 3 1 4 Base 11: (A, B, etc. represent 10, 11, etc..) A, B, C: 18 solutions. From now on, I'll only show the multiple matrix ones. A: 5 A 1 0 9 2 6 7 4 2 3 8 0 9 2 ==> 6 8 3 2 3 8 ==> 9 0 1 6 8 3 1 7 4 9 0 1 4 7 5 B: 9 3 4 5 A 1 7 6 8 ==> 0 9 2 5 A 1 6 8 3 C: 8 9 1 2 3 4 6 7 2 ==> 0 1 5 4 5 3 8 A 6 (Note that the B solution ends in an A solution matrix!) Base 12: 19 solutions A: 7 3 4 2 6 8 2 6 8 ==> 9 A 0 9 A 0 5 1 4 B: None C: 3 5 7 1 A 4 6 B 2 ==> 5 3 B A 4 9 8 9 6 Base 13: 71 solutions...and it rapidly increases from here.... The number of solutions rises rapidly, as we might expect, as the more possible values for digits there are in the base, the more likely the set of 9 will be distinct. If we look at solutions which only involve the digits 1-9, then the following is a list of all solutions (for all bases): Base 10: 1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1 Base 11: 7 8 3 8 4 6 8 9 1 4 5 6 5 9 1 6 7 2 1 2 9 3 2 7 4 5 3 (Tested all cases until base 17. After that, no solution can involve a carry. But there are no solutions without carries. So, no more solutions.) I hope this is of some interest. Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. ------------------------------------------------------------------------------- ------------------------- > Ref: Your note of Mon, 19 Oct 92 22:24:47 EST > > Where are you from? Whoops, knew I forgot to put something in. I'm currently a student at the University of Sydney (Australia), doing Computer Science (Honours). Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. ------------------------------------------------------------------------------- ------------------------- By the way, I tried searching for analogous solutions for other sizes and other bases. So the new problems become: Consider an n by n matrix containing the 'digits' from 1 to n^2 in a base b, where b > n^2. The i'th row of the matrix consists of the last n 'digits' of i*(first row). The versions differ in how succeeding matrices may be constructed. Let f be the first row. Version A: The next matrix has rows with 2*f, 3*f, ... , (n+1)*f The j'th matrix has rows j*f, (j+1)*f, ... , (n+1-j)*f Version B: The next matrix has rows with n*f, 2*n*f, ... , n*n*f The j'th matrix has rows (n^(j-1))*f, 2*(n^(j-1))*f, .... , (n^j)*f Version C: The next matrix has rows with (n+1)*f, (n+2)*f, ... , 2*n*f The j'th matrix has rows (1+n*(j-1))*f, (2+n*(j-1))*f, ..., j*n*f Basically these are analogies of the three versions I wrote to you about before. The results I get are: n: 1 base: any above 1 A, B, C: 1 n: 2 base: 5 A, B, C: 3 2 4 1 1 4 3 2 In case B, the second one extends: 4 1 ==> 3 2 3 2 1 4 In case C, the second one also extends: 4 1 ==> 2 3 3 2 1 4 base: 6 A, B, C: 1 4 3 4 3 2 1 2 Note that the only solution to the first problem (no overflow allowed) is 1 4 (in base 6) 3 2 n: 3 base: 10 A, B, C: 1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1 base: 11 A, B, C: 7 8 3 8 4 6 8 9 1 4 5 6 5 9 1 6 7 2 1 2 9 3 2 7 4 5 3 Note the base 10 solutions all solve the first problem, while none of the base 11 solutions do, and there is no second matrix for any of them. n: 4 base: 18 A, B, C: 1 15 14 4 1 15 16 2 2 1 15 16 2 3 13 16 3 13 10 8 3 13 14 4 4 3 13 14 4 7 9 14 5 11 6 12 5 11 12 6 6 5 11 12 6 11 5 12 7 9 2 16 7 9 10 8 8 7 9 10 8 15 1 10 3 13 14 4 3 13 16 2 4 1 15 14 4 3 13 14 7 9 10 8 7 9 14 4 8 3 13 10 8 7 9 10 11 5 6 12 11 5 12 6 12 5 11 6 12 11 5 6 15 1 2 16 15 1 10 8 16 7 9 2 16 15 1 2 In case C, two of them extend: 1 15 16 2 9 7 8 10 2 1 15 16 10 9 7 8 3 13 14 4 ==> 11 5 6 12 4 3 13 14 ==> 12 11 5 6 5 11 12 6 13 3 4 14 6 5 11 12 14 13 3 4 7 9 10 8 15 1 2 16 8 7 9 10 16 15 1 2 Note all of these solutions solve the first problem (no overflow). Unfortunately, my algorithm is O((n!)^2), so any results for n = 5 are not going to be forthcoming soon. Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. ------------------------------------------------------------------------------- ==> pickover/pickover.08.p <== Title: Cliff Puzzle 8: Squares and Squares and Squares .... From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * 1. What is the smallest square with leading digit 1 which remains a square when the leading 1 is replaced by a 2? In other words, if x**2 = 1.........., is there a y**2 = 2......... ? 2. What is the smallest square with leading digit 1 which remains a square when the leading 1 is replaced by a 2 and also remains a square when the leading digit is replaced by a 3? 3. What is the smallest square with leading digit 1 which remains a square when the leading 1 is replaced by a 2, and also remains a square when the leading digit is replaced by a 3, and also remains a square when the leading digit is replaced by a 4? ==> pickover/pickover.08.s <== ------------------------- > 1. What is the smallest square with leading digit 1 which remains a > square when the leading 1 is replaced by a 2? > 11025 ( 105 * 105 ) ---- 21025 ( 145 * 145 ) > > 2. What is the smallest square with leading digit 1 which remains a > square when the leading 1 is replaced by a 2 and also remains a square > when the leading digit is replaced by a 3? > No solution till 1,000,000,000. > 3. What is the smallest square with leading digit 1 which remains a > square when the leading 1 is replaced by a 2, and also remains a square > when the leading digit is replaced by a 3, and also remains a square > when the leading digit is replaced by a 4? > > No solution till 1,000,000,000. The property that you are looking for ( however with different leading digits ) is owned by the following numbers. 2025 3025 ------------- 11025 21025 57600 67600 --------------- 202500 302500 342225 442225 ------------------ 1102500 2102500 3515625 4515625 5760000 6760000 ------------------- 11390625 21390625 20250000 30250000 34222500 44222500 ---------------------- 110250000 210250000 196700625 296700625 351562500 451562500 576000000 676000000 ------------------------- This is probably of no use to you, but, anyway. ------------------------- In article <1992Oct20.184149.51596@watson.ibm.com> you write: >Title: Cliff Puzzle 8: Squares and Squares and Squares .... >1. What is the smallest square with leading digit 1 which remains a >square when the leading 1 is replaced by a 2? >In other words, if x**2 = 1.........., is there a y**2 = 2......... ? (Isn't this first part an old puzzle?) 105^2=11025; 145^2=21025. In general we want 10^k=(y-x)(y+x) and 1.5 < (y/x)^2 < 2. Thus y+x and y-x must be factors of 10^k of the same parity whose ratio is between 5.828... and 9.899... (these are (t+1)/(t-1) for t^2=2 and 1.5 respectively). The smallest solution (x,y)=(105,145) corresponds to the factorization 10^4=40*250; gp/pari's "fordiv" function allows one to easily list all primitive solutions [i.e. not obtained from a smaller solution by multiplying x,y by the same power of 10] with x^2 and y^2 each having at most (say) 50 digits: [x,y]= [145, 105] [17225, 14025] [454625, 326625] [53948125, 43708125] [1425503125, 1015903125] [168971890625, 136203890625] [529265958203125, 424408358203125] [1657888279384765625, 1322343959384765625] [5193483785077392578125, 4119741961077392578125] In fact it can be seen that the primitive solutions correspond to integer linear combinations of log(2) and log(5) lying in a certain fixed interval (determined by the bounds 5.828... and 9.899...), which probably explains the regular growth of this list. >2. What is the smallest square with leading digit 1 which remains a >square when the leading 1 is replaced by a 2 and also remains a square >when the leading digit is replaced by a 3? There is no such beast, since the three squares would constitute an arithmetic progression of integer squares of common difference 10^k, and so give an A.P. of 3 rational squares of common difference 1 or 10 --- which is known to be impossible by a "2-descent" argument (the case of common difference 1 is already due to Fermat). [We were lucky here: in a different number system this argument might fail; for instance the squares of 7/2, 17/2, 23/2 are an A.P. of common difference 60, the sexagesimal base. (Some numerology: 7,17,23 are the first three primes of which 2 is a quadratic residue.) Still, given the base b, the general theory of elliptic curves indicates that the rational solutions of Y^2-X^2=Z^2-X^2=b are rather sparsely distributed (the number of d-digit solutions growing as some power of d), and the extra condition that they arise by changing only the initial digits of three integer squares is strong enough to ensure that there are at most finitely many solutions; with yet more powerful methods one can even provably list them all.] >3. What is the smallest square with leading digit 1 which remains a >square when the leading 1 is replaced by a 2, and also remains a square >when the leading digit is replaced by a 3, and also remains a square >when the leading digit is replaced by a 4? Of course the above solution to part 2 also disposes of this part; alternatively I could apeal to another classic result of Fermat: there is no 4-term A.P. of rational squares. My question: why all the blank spaces at the end of every line? --Noam D. Elkies (elkies@zariski.harvard.edu) Dept. of Math., Harvard Univ., Cambridge, MA 02138 ------------------------- I dunno the direct answer to your squares problem. I do know, however, that phi (from the Golden Ratio--approx 0.61), which is defined as the number x such that x + 1 = x^2. It just so happens that phi+1 and (phi+1)^2 differ by only 1. (1.61 and 2.61) The rest of the digits are the SAME! Phi = (Sqrt(5)-1)/2. Phi+1 = (Sqrt(5)+1)/2 phi+2 = (Sqrt(5)+3)/2 (Phi+1)^2= (5+2*Sqrt(5)*1+1)/4 = (2*Sqrt(x)+6)/4 = (Sqrt(x) + 3)/2 Notice how that all works out? Perhaps this property will bring you closer to an answer. I just sent you all my personal data in a previous letter concerning your 123 problem. Let me know what you think of this approach, ok? Thanks in advance! --Joseph Zbiciak im14u2c@camelot.bradley.edu ------------------------- In article <1992Oct20.184149.51596@watson.ibm.com> you write: : 2. What is the smallest square with leading digit 1 which remains a : square when the leading 1 is replaced by a 2 and also remains a square : when the leading digit is replaced by a 3? This is not possible. One of these numbers would leave a remainder of 2 when divided by 3, and no square is congruent to 2 modulo 3. -- David Radcliffe radcliff@csd4.csd.uwm.edu ------------------------- In article <1992Oct20.184149.51596@watson.ibm.com> you write: : 1. What is the smallest square with leading digit 1 which remains a : square when the leading 1 is replaced by a 2? 11025. I found, by hand, all integral solutions to (x+y)(x-y) = 10000. The solution (145,105) is the only one with 10000 < y^2 < 20000. You have permission to use my solution, but not my name. -- David Radcliffe radcliff@csd4.csd.uwm.edu ------------------------- Well, as a previous poster already mentioned on Rec.puzzles, there are only 4 solutions to the initial problem. They are 192, 219, 293, and 327. None of these solutions can be connected to others as in part 2 of your problem. I first extended the problem to allow any multipliers. So the second row must be some multiple of the first and the third some other multiple of the first. I found 19 solutions to this problem. However, there is still no way to chain a second solution to the first. Then I allowed 0s. Now there are 134 solutions. There are also 17 2-chains. There are two 3-chains which I will list here: 192 394 *2= 384 *3=1182 *3= 576 *4=1576 *7=4032 now the same as the other solution. *9=5184 *4= 736 *5= 920 I will be more than happy to send you all 134 solutions if you really want them! I also have Pascal source code. Comments on some of your other problems will follow. Dan Cory Senior, Stanford perm. address: 55 Cedar St. Chapel Hill, NC 27514 school address: PO Box 13113 Stanford, CA 94309 Should you use any of my results, please send a copy of the work to the permanent address above. ------------------------- In article <1992Oct20.184149.51596@watson.ibm.com>, you write: |> Title: Cliff Puzzle 8: Squares and Squares and Squares .... |> 1. What is the smallest square with leading digit 1 which remains a |> square when the leading 1 is replaced by a 2? 11025 = 105^2, 21025 = 145^2. |> 2. What is the smallest square with leading digit 1 which remains a |> square when the leading 1 is replaced by a 2 and also remains a square |> when the leading digit is replaced by a 3? |> |> 3. What is the smallest square with leading digit 1 which remains a |> square when the leading 1 is replaced by a 2, and also remains a square |> when the leading digit is replaced by a 3, and also remains a square |> when the leading digit is replaced by a 4? These two cases never occur. Proof: (This was a LOT harder than I thought it would be when I started!) The original problem can be reduced to: "Find positive integers x,y,n such that y^2-x^2 = 10^n and 10^n < x^2 < 2*10^n." [1] The second problem amounts to finding x,y,z,n which meet the above conditions, plus z^2-y^2=10^n. For the second problem, look at the set of solutions to z^2-y^2 = 10^n, 2*10^n < y^2 < 3*10^n. [2] A solution to the second problem consists of x,y,z,n, where x,y,n solve the original problem and y,z,n solve the above system. The first equation in [1] can be factored into (y-x)(y+x) = 10^n = 2^n * 5^n. Similarly (z-y)(z+y) = 10^n. Since x,y,z are integers, we must have y+x = 2^a * 5^b, y-x = 2^(n-a) * 5^(n-b) z+y = 2^c * 5^d, z-y = 2^(n-c) * 5^(n-d) where a,b,c,d are integers. When a=c and b=d, y+x = z+y and y-x = z-y, which leads to a contradiction. Then 2y = 2^a * 5^b + 2^(n-a) * 5^(n-b) = 2^c * 5^d + 2^(n-c) * 5^(n-d) However, in the last equality above, divide both sides by 2^f, where f is the smallest of a, c, n-a, and n-c. The result is: 2^(a-f) * 5^b + 2^(n-a-f) * 5^(n-b) = 2^(c-f) * 5^d + 2^(n-c-f) * 5^(n-d) [3] Now, at least one of the four products above is a product of only 5's, and is odd. Only one is odd unless a=c, 2a=n, or 2c=n. If a=c, then either b=d (contradiction) or z+y is at least a factor of 5 larger than y+x. However, considering sqrt(3)*sqrt(10^n) < z < 2*sqrt(10^n) sqrt(2)*sqrt(10^n) < y < sqrt(3)*sqrt(10^n) sqrt(10^n) < x < sqrt(2)*sqrt(10^n) we have: (sqrt(3)+sqrt(2))*sqrt(10^n) < z+y < (2+sqrt(3))*sqrt(10^n) (1+sqrt(2))*sqrt(10^n) < y+x < (sqrt(3)+sqrt(2))*sqrt(10^n) and then (z+y)/(y+x) < (2+sqrt(3))/(1+sqrt(2)) < 5. If a exactly equals n/2: In the case that b=a=n/2, y+x = y-x, so x=0 (not possible). If b<n/2, y-x>y+x, but we want x to be positive, so b>n/2. Since b and n/2 are integers (remember n/2=a), b-(n-b) >= 2, and (y+x)/(y-x) >= 25. This gives (y+x) >= 25(y-x), (y+x+y-x) = 2y >= 26(y-x), y >= 13y-13x, 13x >= 12y, x/y >= 12/13 x^2/y^2 >= 144/169 However, we know 10^n < x^2 < 2*10^n, and y^2 = x^2 + 10^n, so x^2/y^2 varies between 1/2 and 2/3, and cannot be greater than 144/169. Similarly, when c=n/2, the same argument applies, and in the final step we know y^2/z^2 varies between 2/3 and 3/4. Finally, we've eliminated all cases where more than one of the terms in [3] is odd. With exactly one term odd, we have odd=even, a contradiction, so there is no solution. -- ----w-w--------------Joseph De Vincentis--jwd2@owlnet.rice.edu---------------- ( ^ ) Disclaimer: My opinions do not represent those of Owlnet. (O O) Owlnet: George R. Brown School of Engineering Educational Network. v-v (Unauthorized use is prohibited.) (Being uwop-ap!sdn is allowed.) ------------------------- G'day Cliff! > * * * > > 1. What is the smallest square with leading digit 1 which remains a > square when the leading 1 is replaced by a 2? > > In other words, if x**2 = 1.........., is there a y**2 = 2......... ? The smallest I could find was 105**2 = 11025 145**2 = 21025 Indeed, an exhaustive search shows that this is the smallest. The other pairs I found (after a few minutes playing with pen and paper - I could probably write a program to generate them ad nauseum, but I've got a draft thesis to write...) were: 3375**2 = 11390625, 4625**2 = 21390625 14025**2 = 196700625, 17225**2 = 296700625 326625**2 = 106683890625, 454625**2 = 206683890625 I don't know what pattern there is in them. Of course, if x is a solution, then so is 10*x. So these give solutions for 1050*1050 = 1102500, etc. > 2. What is the smallest square with leading digit 1 which remains a > square when the leading 1 is replaced by a 2 and also remains a square > when the leading digit is replaced by a 3? > > 3. What is the smallest square with leading digit 1 which remains a > square when the leading 1 is replaced by a 2, and also remains a square > when the leading digit is replaced by a 3, and also remains a square > when the leading digit is replaced by a 4? I'll answer part 3 first. If such a square exists, then observe that we have 4 squares in arithmetic progression (common difference a power of 10). There is a well known theorem that there is no set of four squares in arithmetic progression, so there is no solution to part 3. Now, for part 2. We have 3 squares in arithmetic progression. Another well known (and not too hard to derive) theorem states that for three squares in arithmetic progression, their common difference is of the form: D = 4 * K^2 * m * n * (m^2 - n^2) = 4 * K^2 * m * n * (m + n) * (m - n) Now, this value is a power of 10. So the only primes in its factorisation are 2 and 5. Hence neither m nor n is divisible by 3. So (m^2 - n^2) is divisible by 3. Hence a power of 10 is divisible by 3. Contradiction. So now such set of three squares exist (which also proves part 3). Cheers, Geoff. PS: I assume you still have whatever details of mine you care about. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. ------------------------------------------------------------------------------- ------------------------- Here is the solution I just posted to rec.puzzles. Note that I changed my mind on this puzzle! Dan Cory Senior, Stanford PO Box 13113 Stanford, CA 94309 ypay@leland.stanford.edu Newsgroups: rec.puzzles Subject: Re: Cliff Puzzle 8: Squares and Squares ... (SPOILER) Approved: news-answers-request@MIT.Edu Summary: solutions to part 1, no solutions to parts 2 or 3 Expires: References: <1992Oct20.184149.51596@watson.ibm.com> Sender: Followup-To: Distribution: Organization: DSG, Stanford University, CA 94305, USA Keywords: squares, cliff, 8, gcd In article <1992Oct20.184149.51596@watson.ibm.com> cliff@watson.ibm.com (cliff) >1. What is the smallest square with leading digit 1 which remains a >square when the leading 1 is replaced by a 2? >In other words, if x**2 = 1.........., is there a y**2 = 2......... ? We write this condition as the following equations with x,y,a integers: y^2-x^2=10^a 1*10^a<=x^2<=2*10^a 2*10^a<=y^2<=3*10^a We factor the first equation: (y-x)(y+x)=10^a. Let u=x+y. Then 10^a/u=x-y. Since x+y>x-y, u>10^a/u so u>10^(a/2) Then x=(u-10^a/u)/2 and y=(u+10^a/u)/2. Subsitute these equations into the inequalities above. For x we get: 10^a<=((u-10^a/u)/2)^2<=2*10^a Take the square root of both (all three?) sides: 10^(a/2)<=(u-10^a/u)/2<=sqrt(2)*10^(a/2) Multiply through by 2 and divide through by 10^(a/2). 2<=u/10^(a/2)-10^(a/2)/u<=2*sqrt(2) Let v=u/10^(a/2). So v>1. Then: 2<=v-1/v<=2*sqrt(2). We solve these two inequalities. First the left: v-1/v>=2 v^2-2v-1>=0 v>=(1+sqrt(2)) or v<=(1-sqrt(2)). v-1/v<=2*sqrt(2) v>=(sqrt(2)+sqrt(3)) or v<=(sqrt(2)-sqrt(3)). Since v>1, we drop the negative solutions and find: 1+sqrt(2) <= v <= sqrt(2)+sqrt(3). or 1+sqrt(2) <= u/10^(a/2) <= sqrt(2)+sqrt(3). We can do the same for y but we will find the same restriction on u. Now we remember that u|10^a (u divides 10^a). Therefore u must be a power of 2 times a power of 5. Let u=5^b*2^c with b,c integers less than or equal to a. Since we are going to divide it by 2, we must have c>=1. Then we need to find a,b,c such that: 1+sqrt(2) <= 5^b*2^c/10^(a/2) <= sqrt(2)+sqrt(3) These will give us u which will in turn determine x and y. So take the log base 10 of all three sides. Since log is increasing, we do not change the direction of inequality. Thus: log(1+sqrt(2)) <= b*log(5)+c*log(2)-a/2 <= log(sqrt(2)+sqrt(3)) Multiply through by 2: 2*log(1+sqrt(2)) <= 2*b*log(5)+2*c*log(2)-a <= 2*log(sqrt(2)+sqrt(3)) If we approximate log(5) and log(2), this is sort of a Diophantine equation. Since log(5) is very very close to 0.7 and log(2) is very very close to 0.3, our approximations will be okay to find low solutions. If we want big solutions then we need to use better convergents. We can calculate the boundary logs as accurately as necessary. So: 0.77 <= 7/5*b+3/5*c-a <= 0.99 Multiply through by 5: 3.8 <= 7*b+3*c-5*a <= 4.9 So we must find a,b,c such that 7*b+3*c-5*a = 4, with a>b>=0 and a>=c>0. There are many good ways to solve this but we will just pick a small solution. b=3, c=1, a=4 (7*3+3-5*4=21+3-20=4) Then u=5^3*2^1=250. So y+x=250 and y-x=10^a/u=10^4/250=40. Then y=145 and x=105. y^2=21025 and x^2=11025. This is, in fact, the smallest solution (it is easy to show that there is no solution to the 7*b+3*c-5*a with a<4 and a>b>=0,a>=c>0). >2. What is the smallest square with leading digit 1 which remains a >square when the leading 1 is replaced by a 2 and also remains a square >when the leading digit is replaced by a 3? We note from above that y=(5^b*2^c+10^a/(5^b*2^c)/2 or 2y=5^b*2^c+5^(a-b)*2^(a-c). Should we now repeat the problem for a square with leading digit 2 that is replaced by a 3, everything is the same except that y is now the smaller of the pair. Thus: 2y=5^B*2^C-5^(a-B)*2^(a-C) where B and C are different from b and c above but a is necessarily the same (since we want the difference to be the same power of 10 for each transition). Combining the two we get: 5^b*5^c+5^(a-b)*2^(a-b)=5^B*2^C-5^(a-B)*2^(a-C). The proof that this has no solutions is too small to fit in the margin of this posting. >3. What is the smallest square with leading digit 1 which remains a >square when the leading 1 is replaced by a 2, and also remains a square >when the leading digit is replaced by a 3, and also remains a square >when the leading digit is replaced by a 4? There is no solution since there is no solution to part 2. Dan Cory ==> pickover/pickover.09.p <== Title: Cliff Puzzle 9: 3-Atoms and Growth From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Start with 3 digits: 1, 2, and 3. Each succeding row repeats the previous three rows, in order, as you can see from the following diagram. 1 2 3 123 23123 312323123 12323123312323123 2312331232312312323123312323123 1. What is the sum of digits in the 100th row? 2. Get rid of all the twos. Here I've replaced each of them with a "." 1 . 3 1.3 .31.3 31.3.31.3 1.3.31.331.3.31.3 .31.331.3.31.31.3.31.331.3.31.3 In the last row of this diagram, there are three different species: 31, 331 and 3. How many different species are there in row 30? 3. When the sequence first hits a three, it now undergoes an enzymatic cleavage, and the digits on the right of the 3 are swapped with the digits on the left. 1 2 3 123 23123 now becomes 12323 312312323 now becomes 123123233 Now answer the question posed in question 2. ==> pickover/pickover.09.s <== ------------------------- Subject: Re: Cliff Puzzle 9: 3-Atoms and Growth (PARTIAL SPOILER) Newsgroups: rec.puzzles References: <1992Oct20.184304.37364@watson.ibm.com> In article <1992Oct20.184304.37364@watson.ibm.com>, Cliff Pickover writes: > Start with 3 digits: 1, 2, and 3. > Each succeding row repeats the previous three rows, in order > as you can see from the following diagram. > 1 > 2 > 3 > 123 > 1. What is the sum of digits in the 100th row? This one's easy. You basically have a Tribonacci sequence with the initial conditions S_1 = 1, S_2 = 2, S_3 = 3 and S_n = S_{n-1} + S_{n-2} + S_{n-3} for n>3. Thus, it's possible to find a closed form of the type c_1*r_1^n + c_2*r_2^n + c_3*r_3^n. Indeed, letting T_i be the standard Tribonnaci sequence which has initial values T_1 = 1, T_2 = 1, T_3 = 1 we can play a little game by noting the T's go 1 1 1 3 5, and so by linearity S_i = ( T_i + T_{i+2} )/2, hence S_100 = ( T_100 + T_102 )/2. ------------------------- Dear Mr. Pickover, I found your "123" problem interesting. Here's the answers that I came up with. (Note: my personal info that you requested that I include is at the end of the document.) > * * * >Start with 3 digits: 1, 2, and 3. >Each succeding row repeats the previous three rows, in order, >as you can see from the following diagram. >1 >2 >3 >123 >23123 >312323123 >12323123312323123 >2312331232312312323123312323123 >1. What is the sum of digits in the 100th row? Define an arithmetic series as follows: (Note: Read a_1 as "a sub 1" and a_(n-1) as "a sub n-1". I have to do this because I can't use subscripts here.) a_1 = 1 a_2 = 2 a_3 = 3 a_n = a_(n-3) + a_(n-2) + a_(n-1); n>=4 The sum of each line is the sum of it's parts, so therefore, the sum of each row is the sum of the previous three rows' sums. a_30 = 45152016 (I wrote a simple basic program to calculate it.) >2. Get rid of all the twos. Here I've replaced each of them with a "." >.31.3 >31.3.31.3 >1.3.31.331.3.31.3 >.31.331.3.31.31.3.31.331.3.31.3 >In the last row of this diagram, there are three different species: 31, >331 and 3. How many different species are there in row 30? First, let me show that no "new" species will develop, other than those seen in the sample few lines above: First, notice that there are four unique species above: "1","3","31","331". Next, notice that the first species on a line goes in cycles of 3. (Remember how we're building successive rows. The first row repeated on a line is the row three back. Hence the repeating pattern.) Also notice that the ends of the rows do not change, this time because the last row represented on the current row is the row directly previous (and hence, it ends the same.) Because we are building successive rows via concatination, then only locations within new rows where "new" species may be found ("new" meaning not seen in any previous rows) is where the ends of two rows meet in the new row. Since we know that the "end" of each row is limited to ".3" and the "beginnings" of each row cycle through "31", "1", ".", the only possible combinations we can make are "331", "31", and "3". Since we alreadly have seen these, it is now obvious that we will create no more new species. Next, let me show what species we WILL see: The species "3" is on the end of every line. Therefore it will be in row 30. The species "31" and the species "331" are both imbedded in a row previous to row 30. Therefore they will be in row 30, because the "middle parts" of each row are duplicated down the list, not modified. The species "1" only shows up every third row. It happens to occur on rows such that (Row #) mod 3 = 1. Because 30 mod 3=0, the species "1" will NOT occur in row 30. Hence, we have the three species "3","31","331" occuring in row 30. >3. When the sequence first hits a three, it now undergoes an enzymatic >cleavage, and the digits on the right of the 3 are swapped with the >digits on the left. >1 >2 >3 >123 >23123 now becomes 12323 >312312323 now becomes 123123233 >Now answer the question posed in question 2. I'm not taking the time to work this one out entirely. It appears that this algorithm forces 1's out in front all of the time, and keeps appending 3's on the end of the row. Hence, you'll see a proliferation of species such as "3331","33331","333331", etc. It also appears that in row 30, you will have all the species from "3" , "31", "331","3331", "33331", etc up to "33333333333333333333333331". Now, I haven't doublechecked my work here... I've been up all night, and am too tired to double check my conjecture here. But, I believe that I am right, or at least on the right track. I hope these answers help you. I have two questions in return: "Are you the 'pickover' responsible for many of the Fractint fractal types?" and "Were my answers above even close?" I apologize if my answers seemed a little rough & non-formal at points. I hope you understand my explanation above. Thanks for the mental workout. I hope that this helps you, once again. Hope to hear from you soon! -- Joseph Zbiciak im14u2c@camelot.bradley.edu Here's that personal data to requested that I include: I am Joseph Zbiciak, an Electrical/Computer Engineering Major at Bradley University, Peoria, IL. My current address is as follows: Room 121, Heitz Hall 912 N Elmwood, Peoria, IL 61606 My e-mail address is im14u2c@camelot.bradley.edu. Other info: Year in school: Freshman, DOB: 08/29/75 Academic standing: good Favorite toy: his computer Favorite hobby: spelunking through the internet looking for tidbits like this question here. If you need any more information, let me know. Note: I did not post this on the nn yet. Feel free to for me, however. Thanks! -- ------------------------- |> 3.When the sequence first hits a three, it now undergoes an enzymatic |> cleavage, and the digits on the right of the 3 are swapped with the |> digits on the left. |> |> 1 |> 2 |> 3 |> 123 |> 23123 now becomes 12323 |> 312312323 now becomes 123123233 >From how I understand the descriptive rule I get: 1 2 3 123 becomes 312 23123 becomes 12332 331223123 becomes 312231233 >From your example it seems that the trailing 3 is not regarded as a 'first' 3 (123 is not changed), nor is it regarded as a digit to be swapped (as in the two other examples). Is this how the rule should be interpreted? And ... Keep up the good work, these are really good puzzles!! -- stein.kulseth@nta.no (Norwegian Telecom Research) 'When murders are committed by mathematics, they can be solved by mathematics. Most of them aren't, and this one wasn't' - Nick Charles (Dashiell Hammett's "The Thin Man") ------------------------- Dear Dr. Pickover, I found your "123" problem interesting. Here's the answers that I came up with. (Note: my personal info that you requested that I include is at the end of the document.) > * * * >Start with 3 digits: 1, 2, and 3. >Each succeding row repeats the previous three rows, in order, >as you can see from the following diagram. >1 >2 >3 >123 >23123 >312323123 >12323123312323123 >2312331232312312323123312323123 >1. What is the sum of digits in the 100th row? Define an arithmetic series as follows: (Note: Read a_1 as "a sub 1" and a_(n-1) as "a sub n-1". I have to do this because I can't use subscripts here.) a_1 = 1 a_2 = 2 a_3 = 3 a_n = a_(n-3) + a_(n-2) + a_(n-1); n>=4 The sum of each line is the sum of it's parts, so therefore, the sum of each row is the sum of the previous three rows' sums. a_30 = 45152016 (I wrote a simple basic program to calculate it.) >2. Get rid of all the twos. Here I've replaced each of them with a "." >.31.3 >31.3.31.3 >1.3.31.331.3.31.3 >.31.331.3.31.31.3.31.331.3.31.3 >In the last row of this diagram, there are three different species: 31, >331 and 3. How many different species are there in row 30? First, let me show that no "new" species will develop, other than those seen in the sample few lines above: First, notice that there are four unique species above: "1","3","31","331". Next, notice that the first species on a line goes in cycles of 3. (Remember how we're building successive rows. The first row repeated on a line is the row three back. Hence the repeating pattern.) Also notice that the ends of the rows do not change, this time because the last row represented on the current row is the row directly previous (and hence, it ends the same.) Because we are building successive rows via concatination, then only locations within new rows where "new" species may be found ("new" meaning not seen in any previous rows) is where the ends of two rows meet in the new row. Since we know that the "end" of each row is limited to ".3" and the "beginnings" of each row cycle through "31", "1", ".", the only possible combinations we can make are "331", "31", and "3". Since we alreadly have seen these, it is now obvious that we will create no more new species. Next, let me show what species we WILL see: The species "3" is on the end of every line. Therefore it will be in row 30. The species "31" and the species "331" are both imbedded in a row previous to row 30. Therefore they will be in row 30, because the "middle parts" of each row are duplicated down the list, not modified. The species "1" only shows up every third row. It happens to occur on rows such that (Row #) mod 3 = 1. Because 30 mod 3=0, the species "1" will NOT occur in row 30. Hence, we have the three species "3","31","331" occuring in row 30. >3. When the sequence first hits a three, it now undergoes an enzymatic >cleavage, and the digits on the right of the 3 are swapped with the >digits on the left. >1 >2 >3 >123 >23123 now becomes 12323 >312312323 now becomes 123123233 >Now answer the question posed in question 2. I'm not taking the time to work this one out entirely. It appears that this algorithm forces 1's out in front all of the time, and keeps appending 3's on the end of the row. Hence, you'll see a proliferation of species such as "3331","33331","333331", etc. It also appears that in row 30, you will have all the species from "3" , "31", "331","3331", "33331", etc up to "33333333333333333333333331". Now, I haven't doublechecked my work here... I've been up all night, and am too tired to double check my conjecture here. But, I believe that I am right, or at least on the right track. Thanks for the mental workout. I anxiously await more such puzzles! Hope to hear from you soon! -- Joseph Zbiciak im14u2c@camelot.bradley.edu Here's that personal data to requested that I include: I am Joseph Zbiciak, an Electrical/Computer Engineering Major at Bradley University, Peoria, IL. My current address is as follows: Room 121, Heitz Hall B 912 N Elmwood, Peoria, IL 61606 My e-mail address is im14u2c@camelot.bradley.edu. Other info: Year in school: Freshman, DOB: 08/29/75 Academic standing: good Favorite toy: his computer Favorite hobby: spelunking through the internet looking for tidbits like this question here. ==> pickover/pickover.10.p <== Title: Cliff Puzzle 10: The Ark Series From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * 1. Given a large ark containing 2 individuals of every animal species in the world, what would be the approximate total weight of all the organisms? How would your answer differ if you included every plant, bacterial, and fungal organism? 2. Assume that all other organisms on earth were dead except for those on the ark in question 1, and that the animals were released 1000 years ago. What would you expect to be surviving today? (Assume that, where applicable, a male and female were used for each species.) 3. Assume that the year is 1992 and that it rained for 40 days, and the rain covered all the land on the earth. Further assume that the flood waters receded to pre-flood days within several months. What would be the geopolitical changes as a result of the temporary flood? What would be the ecological changes as a result of the temporary flood? ==> pickover/pickover.10.s <== ------------------------- In article <1992Oct20.184354.165170@watson.ibm.com> you write: |> Title: Cliff Puzzle 10: The Ark Series |> From: cliff@watson.ibm.com |> [ lotsa lines deleted ] |> |> 2. Assume that all other organisms on earth were dead except for those |> on the ark in question 1, and that the animals were released 1000 years |> ago. What would you expect to be surviving today? (Assume that, where ^^^^^^^ |> applicable, a male and female were used for each species.) Were you thinking of parthenogenesis or something ??? |> |> 3. Assume that the year is 1992 and that it rained for 40 days, and the |> rain covered all the land on the earth. Further assume that the flood |> waters receded to pre-flood days within several months. |> |> What would be the geopolitical changes as a result of the |> temporary flood? Dunno about this but it's a safe bet that the Netherlands _wouldn't_ get flooded We've been blocking the sea out for hundreds of years, so we've more experience at it than anyone else. |> |> What would be the ecological changes as a result of the |> temporary flood? Andy. Just my opinions, nobody else's, especially not Oracle's ------------------------- > 1. Given a large ark containing 2 individuals of every animal species > in the world, what would be the approximate total weight of all the > organisms? How would your answer differ if you included every plant, > bacterial, and fungal organism? 1000 tons (guessed 10 million species with an average weight of 100 grams, insects push this number down with their huge number of species). No increase through bacteriae or fungi, but maybe with plants. (You were unspecific: All living species?) > 2. Assume that all other organisms on earth were dead except for those > on the ark in question 1, and that the animals were released 1000 years > ago. What would you expect to be surviving today? (Assume that, where > applicable, a male and female were used for each species.) None. I think it's common knowledge with biologists that you need at least ~50 individuals of a species to keep genetic health --- aside from the problem of both a male and female baby surviving. > 3. Assume that the year is 1992 and that it rained for 40 days, and the > rain covered all the land on the earth. Further assume that the flood > waters receded to pre-flood days within several months. "Covers the land." How deep? To cover *all* land (Himalaya) evenly, you need a depth of 9000 m in most regions, so the question is, how fast will it rise? Do we just have time to put some tins in the boat? Most people don't have one. Most airplanes cannot land but maybe some of them swim. One has to calculate the distribution of swimming things in usual locations. For if people have to swim 500-1000 m in cold water to a beam, most will drown. > What would be the geopolitical changes as a result of the > temporary flood? With the survival of at most 1 percent of the population there will be a completely new beginning. Don't know if they would make the same mistakes, though. Technology will be thrown back, and science more than that. Niven/Pournelle's "Lucifer's Hammer" is an accurate description. > What would be the ecological changes as a result of the > temporary flood? Lack of most animals, especially those dependent of plants (many of them can't live without a day of food). Most plants will grow again after some time. --ralf ************************************************************************ After some tests, I decided to put 4 lines of sig here, because I really like the optical effect. Now there's the problem what to write in it... ************************************************************************ ==> pickover/pickover.11.p <== Title: Cliff Puzzle 11: The Leviathan Number From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Many interesting observations have recently been published concerning various number theory properties of the "number of the beast", 666. In this new puzzle here I ask you to consider the monstrous "leviathan number", a number so large as to make the number of electrons, protons, and neutrons in the universe (10**79) pale in comparison. (It also makes a googol (10**100) look kind of small). The leviathan number is defined as (10**666)!, where the "!" indicates factorial. 1. What are the first 6 digits of the leviathan number? Hint: you need not actually compute the leviathan to determine this. If you can determine the first 6 digits, please carefully spell out your method. 2. Could modern supercomputers compute the leviathan, or will this beyond the realm of humankind for the next century? 3. Even if we cannot compute the leviathan, how many other characteristics of this number can we write down. ==> pickover/pickover.11.s <== ------------------------- Subject: Re: Cliff Puzzle 11: The Leviathan Number (PARTIAL SPOILER) Newsgroups: rec.puzzles References: <1992Oct21.135208.119425@watson.ibm.com> In article <1992Oct21.135208.119425@watson.ibm.com>, Cliff Pickover writes: > The leviathan number is defined as (10**666)!, where the "!" indicates > factorial. > 1. What are the first 6 digits of the leviathan number? The simplest technique would be to use Stirling's formula to compute the mantissa, i.e. frac( log(n) ) = frac( log(2*pi)/2 + log(n)/2 n*(log(n)-log(e)) ). In our case n = 10^666, so this equals frac( log(2*pi)/2 + 333 + 10^666*(666-log(e)) ) = frac( log(2*pi)/2 + 10^666*(1-log(e)) ), so we'd basically need to know something like 10 digits to the right of the decimal point for log(2*pi)/2, and something like 700 digits for log(e) (which is easily doable). We then compute (1-log(e)), shift the digits 666 spaces to the left, and we're all set. > 2. Could modern supercomputers compute the leviathan, or will this > beyond the realm of humankind for the next century? The number of digits is more than 10^668, and this compares unfavorably to the number of particles in the universe. Furthermore, even if a googol digits could be output per second, you'd never make it before the end of the universe. So, I'd say it's beyond the realm of humanity, period. > 3. Even if we cannot compute the leviathan, how many other > characteristics of this number can we write down. As another puzzle, how many zeroes does it end with, and what are the last two non-zero digits? .qq &EXIT THIS FILE HAS BEEN RECEIVED FROM BITNET The file may be executable. Before removing this header you must understand what the code will do. You must also have the appropriate intellectual property agreements in place before receiving the code into IBM. If you have any questions, contact your manager. The contents of the file has been shifted right by one character. Filename=(none) Filetype=(none) RECFM=F LRECL=80 Records=21 The file received from the BITNET gateway begins below the next line. ------------------------------------------------------------------------ Date: Thu, 22 Oct 1992 07:12 EDT From: <FRAMEM@UNION> Subject: RE: googol! To: CLIFF@YKTVMV Original_To: Jnet%"CLIFF@YKTVMV" Hi, Cliff. The log10(e) comes from applying Stirling's approximation for the factorial: for large n, n! is approximately sqrt(2*pi*n)*((n/e)^n). Substitute googol for n, take log10 of both sides, and recall the mantissa of the log10 gives the digits of the original number. In these days of fast symbolic packages allowing exact computation of large factorials (though presumably not so large as a googol), people forget Stirling's formula. Until a few years ago, this was the only way to find factorials (albeit, only approximately) for large numbers. 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