Really complicated code

[jebby]

Hey guys,

I can't seem to break this code. Can you guys help me? I only worked out 3 is 0 and 5 is either 4 or 9 and 1 is either 2 or 3.

The equations below are written in code such that each digit shown represents some other digit. Break the code, given that each of the following is true in ordinary base ten arithmetic:

8+7=62 5+3=5 12+8=23
50+9=54 11x1=55 0-9=1

give some indication of how you got your first three or four digits

Thanks guys

 

[Oaky]

Do you know if the code is restricted to Natural numbers or Integers? (ie are negatives or fractions allowed?)

I'm confused, I don't know how you got 5 is 4 or 9, seeing as the 5 and the 3 are only found in the one equation you don't have any basis on which to determine the value of 5.

 

[soroban]

Hello, jebby!

I can't seem to break this code. Can you guys help me?
I only worked out 3 is 0 and 5 is either 4 or 9 and 1 is either 2 or 3.

The equations below are written in code such that each digit shown represents some other digit.
Break the code, given that each of the following is true in ordinary base ten arithmetic:

. . 8 + 7 = 62 . . 5 + 3 = 5 . . 12 + 8 = 23

. . 50 + 9 = 54 . . 11 x 1 = 55 . . 0 - 9 = 1

First, I replaced the digits with letters, forming a standard alphametic.

. . \(\displaystyle \begin{array}{c} \;\;(1) \\ \quad\; A \\ +\;B \\ \hline C\ \end{array}\) . . . \(\displaystyle \begin{array}{c} \;\;(2) \\ \quad\;E \\ +\;F \\ \hline \quad E \end{array}\) . . . \(\displaystyle \begin{array}{c} \;\;(3) \\ G\;D \\ + \;A \\ \hline D\;F \end{array}\) . . . \(\displaystyle \begin{array}{c}\;\;(4) \\ E\;H \\ + \;I \\ \hline E\;J \end{array}\) . . . \(\displaystyle \begin{array}{c}\;\;(5) \\ G\;G \\ \times\;G \\ \hline E\;E \end{array}\) . . . \(\displaystyle \begin{array}{c}\;\;(6) \\ \quad\;I \\ +\;G \\ \hline \quad H \end{array}\)


There are many clues woven in these equations . . .

From (1), \(\displaystyle \boxed{C = 1}\)
. . Also: .\(\displaystyle A + B \;\to\;^1D \;\hdots\;A + B\text{ ends in }D\text{ with "1 to carry"}\)

From (2), \(\displaystyle \boxed{F = 0}\)

From (3), \(\displaystyle D + A \:\to\;^1F \;\hdots\;D+A\text{ ends in }F\text{ with "1 to carry"}\)
. . .Also: .\(\displaystyle D \:=\;G+1\)

From (4), \(\displaystyle H + I \,=\,J\;\hdots H+I\text{ equals }J\text{, no "carry"}\)

From (5), \(\displaystyle G\times G \,=\,E\;\hdots\;G\times G\text{ equals }E\text{, no "carry"}\)
. . Hence, \(\displaystyle G = 2\text{ or }3,\;E = 4\text{ or }9\)

From (6), \(\displaystyle I + G \,=\,H\;\hdots\;I+G \text{ equals }H\text{, no "carry"}\)


Go for it!