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Steffi Problem


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 2<=r<=6, since a ratio of two permutations of the first seven digits must be less than 7654321/1234567=6.2..., and the permutations were stated to be unequal, so r!=1. The case r=3 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 r=6 as a possibility, since a number must be divisible by 3 to be divisible by 6.

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

 (......5)/(......1).
(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

 (7643215)/(2345671)
(2)

But 764321/2345671=3.25843<5, so it is not possible to construct a fraction that is divisible by 5. Therefore, only r=2 and 4 need now be considered.

In general, consider the numbers of pairs of unequal permutations of all the digits 12...k_b in base b (k<b) whose ratio is an integer. Then there is a unique (b=4,k=3) solution

 (312_4)/(123_4)=2,
(3)

a unique (5,4) solution

 (4312_5)/(1234_5)=3,
(4)

three (6,4) solutions

(3124_6)/(1342_6)=2
(5)
(4213_6)/(1243_6)=3
(6)
(4312_6)/(2134_6)=2,
(7)

and so on.

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

bsolutions for digits 12_b, 123_b, ..., 12...(b-1)_b
30
40, 1
50, 0, 1
60, 0, 3, 25
70, 0, 0, 2, 7
80, 0, 0, 0, 68, 623
90, 0, 0, 0, 0, 124, 1183
100, 0, 0, 0, 0, 0, 2338, 24603
110, 0, 0, 0, 0, 0, 3, 598, 5895
120, 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 12345678_(10), there are two that produce three different integers for the same numerator:

(85427136)/(42713568)=2,(85427136)/(21356784)=4,(85427136)/(14237856)=6
(8)
(86314572)/(43157286)=2,(86314572)/(21578643)=4,(86314572)/(14385762)=6.
(9)

Taking the diagonal entries (b,b-1) from this list for b=3, 4, ... gives the sequence 0, 1, 1, 25, 7, 623, 1183, 24603, ... (OEIS A080203).


See also

Divisibility Tests, Pandigital Fraction

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References

Sloane, N. J. A. Sequences A080202 and A080203 in "The On-Line Encyclopedia of Integer Sequences."

Referenced on Wolfram|Alpha

Steffi Problem

Cite this as:

Weisstein, Eric W. "Steffi Problem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/SteffiProblem.html

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