__Cliff Pickover__

An archive of questions and answers that may be of interest to puzzle enthusiasts.

__Question 7 - pickover.07:__

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.Show Answer

> 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. -------------------------------------------------------------------------------

__Question 8 - pickover.08:__

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?Show Answer

> 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 by+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