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)