abc - cba ----- (a,b,c in some order)
Repeat the problem for N digits (4<=N<=10). That is, subtract an N-digit number from its reverse to result in another N-digit number, such that each of the three consist of the same N distinct digits. Try to find the answer with the largest difference.
Source: Original extension of a problem from The Little Giant Encyclopedia of Puzzles.
Most solvers used a computer program, but there were some good insights for logical approaches. Here are some samples of the solutions received.
Here is my own quick solution for N=3:
The middle number of the answer must be either 0 or 9. Since a > c, the middle number must be 9. Since c < a, either b or a is 9. If b is 9, then the first number of the answer has to be c and the last a. Since the center column requires a borrow, 2*c = a-1, AND 2*a = c+10. a has to be odd and < 9, but no solutions result in a valid c. So a is 9. Again 2*c = a-1, so c=4. And the ones column shows b=5.Al Zimmermann lists the solutions:
954 - 459 = 495 7641 - 1467 = 6174 96732 - 23769 = 72963 927801 - 108729 = 819072 9872541 - 1452789 = 8419752 98472501 - 10527489 = 87945012 987654321 - 123456789 = 864197532 9847623501 - 1053267489 = 8794356012 9846732501 - 1052376489 = 8794356012 (tie for maximum difference)Denis Borris shows how many answers there were and the lowest and highest values used in the differences:
N CASES LOW HIGH 3 1 954 954 4 4 2961 9108 5 8 58923 96732 6 56 279351 981207 7 178 2904561 9872541 8 645 23769801 98601372 9 2204 302548671 987654321 10 9568 2348765901 9875316042Philippe Fondanaiche had some excellent insights for limiting the solutions:
N=3 digits Let A=abc and B=cba with a>c D = A - B is of the form 99*(a-c) One solution a - c = 5 ==> A = 954, B = 459, D = 495 N=4 digits A=abcd and B=dcba with a>d D = A - B is of the form 9*[111*(a-d) + 10*(b-c)] N=5 digits D is of the form 99*[101*(a-e) + 10*(b-d)] N=6 digits D is of the form 9*[1111*(a-f) + 1110*(b-e) +100*(c-d)] N=7-10 digits Same approach as above.Since they're short lists, here are the complete answers for N=4 and N=5, from Al Zimmermann:
N=4 4 solutions 9108 - 8019 = 1089 2961 - 1692 = 1269 7641 - 1467 = 6174 5823 - 3285 = 2538 N=5 8 solutions 60273 - 37206 = 23067 60732 - 23706 = 37026 70254 - 45207 = 25047 89604 - 40698 = 48906 76941 - 14967 = 61974 58923 - 32985 = 25938 69723 - 32796 = 36927 96732 - 23769 = 72963
Let us denote by "uvw" the result: abc-cba=uvw where u,v,w are the digits a,b,c, in a suitable order. Of course the problem has a meaning not only in basis 10, but with respect to any basis B>2; let us consider such a generalization. RESULTS If B=10 the unique solution is given by a=9, b=5, c=4. More generally, let Q be any nonnegative integer; when B=6Q+4 or B=6Q+6 there is a unique solution given by a=B-1, b=B/2, c=B/2-1; the same triple also solves the problem when B=6Q+2 (with Q>0), but in such a case a second solution exists: a=(2B-1)/3,b=B-1,c=(B-2)/3. The last triple also works when B has the form B=6Q+5, case where again the solution is unique; in the remaining cases (B=6Q+1, and B=6Q+3) the problem has no solutions. PROOFS No matter which is B one has: 1) a>c; 2) v=B-1; 3) uvw=(B*B-1)*(a-c); in particular u+v+w=0 mod B-1; say (because of point 2, and u,w being digits in basis B) u+w=B-1. Now uvw=(B*B-1)*(a-c) becomes u=a-c+1. The maximum between a,b,c (which coincides with the maximum between u,v,w) is of course the value of v, say B-1. We can then conclude with a simple distinction of cases: if a=B-1 then the known relations imply that the basis B must be even; and this forces b=B/2, c=B/2-1; if a=max[u,w] then the known relations imply that the basis B must be of the form 3X+2; and this forces a=2X+1, b=3X+1, c=X.