A logic problem of numbers

  • Thread starter mhill
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  • #1
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Homework Statement



given a number of four ciphers ABCD and another four cipher number CDAB get the values of A,B,C,D if we know that 2x (ABCD)=CDAB-5

Homework Equations



2X (ABCD)=CDAB-5

The Attempt at a Solution



no idea.. i have try by brute force but got no results only incongruences , for example B=-3 or similar here '2 x' means multiplication
 

Answers and Replies

  • #2
You need to be a bit more elegant about this. Split it into powers of ten... for example in the ABCD cipher, the "A" is 100 times more value than the "A"
 
  • #3
188
1
no way , i get weird responses such us A=c=0 and B=-5/3 how are these kind of problems solved in Number theory ?? thanks.
 
  • #4
HallsofIvy
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Homework Statement



given a number of four ciphers ABCD and another four cipher number CDAB get the values of A,B,C,D if we know that 2x (ABCD)=CDAB-5

Homework Equations



2X (ABCD)=CDAB-5

The Attempt at a Solution



no idea.. i have try by brute force but got no results only incongruences , for example B=-3 or similar here '2 x' means multiplication
Notice that AB and CD always appear together. I would do it this way:

Let X= AB and Y= CD. Then the equation says 2(100X+ Y)= 100Y+ X- 5. Combining the X and Y terms we get the Diophantine equation
-199X+ 98Y= 5. Do you know how to solve that? It has an infinite number of integer solutions but only one that gives a two digit positive integer for both X and Y.
 
  • #5
CompuChip
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This looks quite similar to this post by the way.
 
  • #6
HallsofIvy
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Very similar! That's why I deleted that thread.
 

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