From: utzoo!decvax!harpo!utah-cs!utah-gr!thomas Newsgroups: net.math Title: Re: Rebuttal: Using a computer to solve ABCDE problem (***SPOILER***) Article-I.D.: utah-gr.486 Posted: Sat Jul 17 11:22:46 1982 Received: Sun Jul 18 02:26:57 1982 References: ihuxv.190 Last night as I lay in bed trying to go to sleep, I was thinking about the ABCDE x 4 puzzle and came up with the following observations: ...9*9... x n = ...9*.... (1 <= n <= 9) In other words, a number with a string of 9's in the middle will still have a string of 9's in the middle after multiplication by any digit. If the carry into the string of 9's from the right is n-1, then all the 9's will remain, otherwise the rightmost 9 will be lost. 219*78 x 4 is the only answer? If A...B x n = B...A (2 <= n <= 9) then - A x n (plus possible carry in) = B (no carry out) - B x n = A (plus possible carry out) - If n is even then so is A. This immediately eliminates n=6 or 8. n = 9: A = 1 -> B = 9, with no carry from the second to the left place. There are two ways this can happen: 1. A...B is a string of 1's. This obviously doesn't solve the problem. 2. The second digit is a 0. This leads to a solution: 1 0 8 9 x 9 ------- 9 8 0 1 This solution may be extended by inserting 9's in the middle 1 0 9 9 9 8 9 x 9 ------------- 9 8 9 9 9 0 1 There may be others, I didn't pursue that line too closely. n = 7: A = 1 -> B = 7,8,9; 7*7 = 9, 7*9 = 3, neither = 1. n = 5: A = 1 -> B = 5,6,7,8,9; Obviously 5*something != 1. n = 4: Leads to the previous solution. n = 3: A = 1 -> B = 3,4,5; 3*3 = 9, 3*5 = 5, neither 1. A = 2 -> B = 6,7,8; 3*6 = 8, 3*8 = 4, neither 2. A = 3 -> B = 9; 3*9 = 7 != 3. n = 2: A = 2 -> B = 4,5; 2*4 = 8, 2*5 = 0, neither 2. A = 4 -> B = 8,9; 2*8 = 6, 2*9 = 8, neither 4. QED (Sort of) Note also that any solution may be appended to itself to give a new solution! 21782178 x 4 = 87128712. Unanswered questions: Are there any other solutions for n=9? [ Again, I claim not. ] Are there any solutions for n=4 with the digit 0 included? [ No, a little pencil/paper calculation will show that there is no way to introduce a 0] If repeated digits are not allowed, the two solutions above are clearly the only possible solutions, and if 0 is also not allowed, then the solution for n=4 is the only solution. Note that this is a question which is ONLY answerable by applying some reasoning power. This is true in general of (non-trivial) mathematical questions. A computer brute-force solution may indeed be useful for pointing out a direction of investigation, or for suggesting patterns, but can never directly answer a uniqueness question like this one. =Spencer Thomas (harpo!utah-cs!thomas, thomas@utah-20)