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Arithmetic Code


Date: 05/09/99 at 13:09:00
From: John 
Subject: Breaking the Code

The questions below are written in code such that each digit shown 
represents some other digit. Break the code, given that each of the 
folowing is true in ordinary base ten arithmetic: 8+7 = 62, 5+3 = 5, 
12+8 = 23, 50+9 = 54, 11*1 = 55, 0-9 = 1.

And could you please give me some indication of how you got your 
first three or four digits? 

I don't know where to start. Could you please help me?


Date: 05/10/99 at 11:59:47
From: Doctor Peterson
Subject: Re: Breaking the Code

Hi, John. This is a nice challenge!

The first thing I did was to change the digits to letters to avoid 
confusion. I made a chart like this:

    code digit:  0 1 2 3 4 5 6 7 8 9
    code letter: a b c d e f g h j k
    real digit:  _ _ _ _ _ _ _ _ _ _

Then I rewrote the equations using the letters:

    1: j + h = gc
    2: f + d = f
    3: bc + j = cd
    4: fa + k = fe
    5: bb * b = ff
    6: a - k = b

A couple of these give us some digits almost immediately. In (2), you 
can add d to f and it leaves the f unchanged. What must d be? In (1), 
it's a little more subtle: if you add two single digits and get a two-
digit number, what can the tens digit be?

From here it gets a little harder. I focused on (5); since bb = b*11 
and ff = f*11 (that's why I changed to digits, so I can mix coded and 
real digits), we see that b*11 * b = f*11, so b*b = f. There are only 
four digits that have a one-digit square, and two of those are already 
taken, so there are only two possibilities for b and f. Using each of 
these, I could then find from other equations what c and j are. Then I 
looked at (4) and saw that a + k = e; combining that with (6) I found 
that there were only a few possibilities for k. All but one of these 
conflicted with what I had found for b, f, c, and  j.

That's the main work I did. After that it was just a little work to 
finish.

Don't try to follow every step I described; just take my ideas as 
examples of the sort of thing you can do, and you may find a better 
way to get to the solution.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
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