### 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