A homework problem proposed in Steffi's math class In January 2003 asked students to prove that no ratio of two unequal numbers obtained by permuting all the digits 1, 2, ..., 7 results in an integer. If such a ratio r existed, then some permutation of 1234567 would have to be divisible by r. r can immediately be restricted to
This leaves only the cases r = 2, 4, and 5 to consider. The r = 5 case can be eliminated by noting that in order to be divisible by 5, the last digits of the numerator and denominator must be 5 and 1, respectively
The largest possible ratio that can be obtained will then use the largest possible number in the numerator and the smallest possible in the denominator, namely
But
In general, consider the numbers of pairs of unequal permutations of all the digits in base b (k < b) whose ratio is an integer. Then there is a unique solution
a unique solution
three solutions
and so on.
The number of solutions for the first few bases and numbers of digits k are summarized in the table below (Sloane's A80202).
b | solutions for digits |
3 | 0 |
4 | 0, 1 |
5 | 0, 0, 1 |
6 | 0, 0, 3, 25 |
7 | 0, 0, 0, 2, 7 |
8 | 0, 0, 0, 0, 68, 623 |
9 | 0, 0, 0, 0, 0, 124, 1183 |
10 | 0, 0, 0, 0, 0, 0, 2338, 24603 |
11 | 0, 0, 0, 0, 0, 0, 3, 598, 5895 |
12 | 0, 0, 0, 0, 0, 0, 0, 0, 161947, 2017603 |
As can be seen from the table, in base 10, the only solutions are for the digits 12345678 and 123456789. Of the solutions for
Taking the diagonal entries from this list for b = 3, 4, ... gives the sequence 0, 1, 1, 25, 7, 623, 1183, 24603, ... (Sloane's A80203).
Divisibility Tests, Pandigital Fraction
Sloane, N. J. A. Sequences A80202 and A80203 in "The On-Line Encyclopedia of Integer Sequences." http://www.research.att.com/~njas/sequences/.