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 existed, then some permutation of 1234567 would have to be divisible by . can immediately be restricted to , since a ratio of two permutations of the first seven digits must be less than , and the permutations were stated to be unequal, so . The case can be eliminated by the divisibility test for 3, which says that a number is divisible by 3 iff the sum of its digits is divisible by 3. Since the sum of the digits 1 to 7 is 28, which is not divisible by 3, there is no permutation of these digits that is divisible by 3. This also eliminates as a possibility, since a number must be divisible by 3 to be divisible by 6.

This leaves only the cases , 4, and 5 to consider. The 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 , so it is not possible to construct a fraction that is divisible by 5. Therefore, only and 4 need now be considered.

In general, consider the numbers of pairs of unequal permutations of all the digits in base () whose ratio is an integer. Then there is a unique solution

a unique solution

three solutions

			(5)
			(6)
			(7)

and so on.

The number of solutions for the first few bases and numbers of digits are summarized in the table below (Sloane's A080202).

	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 , there are two that produce three different integers for the same numerator:

			(8)
			(9)

Taking the diagonal entries from this list for , 4, ... gives the sequence 0, 1, 1, 25, 7, 623, 1183, 24603, ... (Sloane's A080203).