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
 |
(1)
|
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
 |
(2)
|
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
 |
(3)
|
a unique 
solution
 |
(4)
|
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 (OEIS 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,
... (OEIS A080203).
