February 2016
The number of possibilities (10!=3,628,800) is well in the range of brute force checking, but we wanted to make it almost doable without a computer.
Here is Hao Zheng's solution
Steps:
1. Using n**i = ban to get the following possible solutions of n, i, b, a
2^9 = 512 (n, i, b, a) = (2, 9, 5, 1) <-- eliminated by step 2
3^5 = 243 (n, i, b, a) = (3, 5, 2, 4)
5^3 = 125 (n, i, b, a) = (5, 3, 1, 2)
5^4 = 625 (n, i, b, a) = (5, 4, 6, 2)
6^3 = 216 (n, i, b, a) = (6, 3, 2, 1) <-- eliminated by step 2
9^3 = 729 (n, i, b, a) = (9, 3, 7, 2)
2. Using a**connect = c (mod inter) to eliminate a = 1, otherwise a = c = 1
3. There are 4 possible solutions for n, i, b, a
(3, 5, 2, 4)
(5, 3, 1, 2)
(5, 4, 6, 2)
(9, 3, 7, 2)
Notice that c and t cannot be 0 because they are the leading digits.
So the number of all possible solutions for
(a, b, c, e, i, m, n, o, r, t)
is 4 * 5 * 4 * 4 * 3 * 2 * 1 = 1920
Write a program to try all the 1920 possibilities, we get the following
(a, b, c, e, i, m, n, o, r, t) =
(2, 6, 8, 0, 4, 9, 5, 1, 7, 3)
But there was another clue: connect = toe (mod 11) (left as an exercise for the reader: David Greer noticed that part). So a**connect = a**toe (mod 2**11-1), which is 2047=89*23, which helps guessing that cm=89 and at=23, leaving a few possibilities to check.
As for the bonus, the error message "XCF MODULE ID FILTER NOT ACCEPTED" is IXC469I and converting 469 to letters, using the solution, yields "IBM".