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Problem Description

I recently received this email:

Dear Delphiforfun,

I have been investigating some variations on a classical problem involving
pandigital fractions and wonder if you have seen solutions.

Classical Problem -- I solved this one
If each of the first 9 letters represents a different number--1 through 9, the
smallest integer represented by: A/BC + D/EF + G/HI = 1; this assumes that the
denominators are two-digit numbers and NOT multiplied.

My Variations -- Need Verification of Solutions
I have modified the classical equation so that the denominators are multiplied
as follows: A/(B*C) + D/(E*F) + G/(H*I) and believe the smallest integral answer
is 2. Can you verify that?

For (A/B)*C + (D/E)*F + (G/H)*I, I think the smallest integral answer is 5.

For (A/B)^C + (D/E)^F + (G/H)^I. the smallest integral answer I have found is
1,100, but I suspect there is a smaller solution.

Regards, Jerry

Well Jerry, assuming that you are not a programmer and were doing this by hand, 1 correct out of 3 isn't bad!



Background & Techniques

I wrote the simple code to solve these 4 problems (the classical version and the three variations)..  I emailed the results back to Jerry,  but so far, have received no response.   Maybe he took it hard that two of his three answers were incorrect. 

Each of the four uses the same technique, namely:

Generate all permutations of the digits 1 through  9.  For each permutation, assign the digits of the result in order to the 9 variables A - I and evalua6te the expression.  We'll do the evaluation in extended mode to retain fractional information.  We check that the result is and integer and if so, that it is a new  minimum before adding it to the potential solutions list.

The GetNext function from our old standby,  the Combo unit, returns the 360,000 or so permutations of the 9 digits.  Any rational numbers whose decimal representation  is an infinite repeating non-zero string cannot be represented exactly in floating point form.   To get around this problem,  we look for values that are very close to an integer (10^-32 in this case).    This test and the new minimum test are incorporated into procedure CheckAndShowValues.  Procedure AssignValues assigns the permutation output to the variables.  So each solution search contains a six statement loop like this, with only the "x:=.." statement changing :

while combos.getnext do
begin
    assignvalues;
    x:=a/(10*b+c)+d/(10*e+f)+g/(10*h+i);  {for example}
    CheckAndShowValues(x);
end;

Running/Exploring the Program 

Suggestions for Further Explorations

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