I'm not certain this is complete/rigorous; It does however lead me to think that this problem could be solved analytically. by those of keener aptitude.
For convenience, let ABCD = N.
1. Divisibility by (10c+d) gives x(10c+d) = 100(10a+b). Let x = 100, so that a = c and b = d. 11b then divides 101(10a+b) and 11a divides 101(10a+b), so a = b, in which case (10b+c) divides N; this is prohibited by the problem.
2. It is given that (10a+b) divides N and hence (10a+b) divides (10c+d), but it has just been shown that (10a+b) cannot equal (10c+d). Hence (10c+d) = n(10a+b), where 1<n<=an<10. Accordingly all the prime factors of (10a+b) are also prime factors of (10c+d), and all the prime factors of (10c+d) are factors either of (10a+b) or n. It follows immediately that all the factors of x are held in common with 100 i.e. {2^k,5^k}
3. (10c+d) is at least 2(10a+b). Since x(10c+d) = 100(10a+b), x cannot exceed 50. (10c+d) is at most 9(10a+b); multipliers that do not evenly divide 100 may be eliminated, giving {x, n} = {20,5:25,2^2:50,2}
4. Let (10a+b) be prime, P. The prime factors of (10c+d) are then, {2,5,P}.
(a) If 5 is a factor of (10c+d) then the problem requires d to be 5 also, compelling n=5, a=1. Since that factor is 'borrowed' from x, x=20, so 20(10c+5) = 1000+100b and c = k+5 b = 2k+1: 1155,1365,1575,1785,1995; Division by (10a+c) quickly rules out all except 1995, which is a valid solution.
(b) Let (10c+d) be even; by parity of reasoning, x= 25, n = 4, or x = 50, n=2.
a = 1;1122,1326,1734,1938,1188,1352,1768,1986 divided by {21,51} produces numerous candidates, but (10a+d) divides none of these.
a = 2; 2346,2958; no candidates
a = 3; 3162,3774; one candidate, 3774, which is a valid solution.
a = 4: 4182,4386,4794; but (10a+d) divides none of these.
5. Let (10a+b) be compound.
(a) Since 2*3*5*7 = 210, (10a+b) has at most 3 prime factors. Since no number less than 100 has two prime factors greater than 10, at least 2 of the factors must be less than 10 {2,3,5,7} of which only 3 and 7 are new.
(b) Further, each of those factors is also a factor of (10c+d), which is at least 2(10a+b), where n = 2 is 'borrowed' from x. But x(10c+d) = 100(10a+b), and every factor of 100 must still be accounted for somewhere. This combination of constraints is only possible if (10a+b) holds NO factors in common with 100.
(c) Next, assume (10a+b) to have some factor greater than 10, say 11. Now (10a+b) = 33, and n=2: 50(10c+d) = 200(33) and 10c+d = 132, which is already outside the permitted range. It follows that not only does (10a+b) have no factors in common with 100, but the sole factors that it can have are 3(singly) and 7 (singly).
(d) Then (10a+b) = 21 gives N = 2142, 2163, 2184. Division by (10b+d) leaves 2184 the sole survivor; it is also a solution.
The probability has already been computed in earlier posts.
Edited on October 21, 2010, 8:30 am

Posted by broll
on 20101021 08:25:14 