lateral thinking - Sum other numbers

Begin with a flagrantly erroneous summation and a woefully vacant substitution table.

           234

         +   5                   Digit        2    3    4    5    6    7    8
        -------           Substitute digit    _    _    _    _    _    _    _
          5678

How can the substitution table be filled out to correct this summation?

This is almost too easy if you just follow these guidelines.

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  Assign 7 unique substitute digits from 0 through 9 for digits 2 through 8 in the table (one digit per digit)

  
  


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  Replace digits in the summation by their substitutes in the table (no other kinds of edits, as the summation and table should be taken at face value)

  
  


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  All numbers and digits are decimal (no notation tricks are involved)

  
  


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  No leading zeros in the total or either summand

  
  
  


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  The summation has a unique solution

  
  

Added:   Regular arithmetic pretty much forces the resultant summation. lateral-thinking allows the guidelines to attain it.

Answer

Making an assumption:-

That if a substitute digit is itself in the lookup table, it will be replaced again.

 Digit               2    3    4    5    6    7    8
 Substitute digit    3    4    9    1    7    8    0 

The Summation becomes:

999 + 1 = 1000 because:
2->3->4->9,
3->4->9,
4->9,
5->1,
5->1,

6->7->8->0,
7->8->0,
8->0

Process:

As the question states, if you follow the guidelines, it should lead you towards the answer

First, as mentioned in the question, there is one possible summation. It must be 999 + 1 = 1000 as a 3 digit number plus a 1 digit number must equal a 4 digit number, and the first digit of the 4 digit number has to be the same as the 1 digit number.

Then, knowing that 6,7,8 must equal 0 we can first assign any one of those digits the substitute digit of zero, lets choose 8.

Since 0 is now used (and the question states the substitute digits must be unique) in order for 6 or 7 to equal 0, the only substitute digit we can assign is 8 (since 8 = 0).

This same logic is then applied to 2,3,4 since they all need to equal 9