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 The magnificent twelve (Posted on 2017-04-07)
Z+E+R+O = 0
O+N+E = 1
T+W+O = 2
T+H+R+E+E = 3
F+O+U+R = 4
F+I+V+E = 5
S+I+X = 6
S+E+V+E+N = 7
E+I+G+H+T = 8
N+I+N+E = 9
T+E+N = 10
E+L+E+V+E+N =11
T+W+E+L+V+E = 12

Solve the above system of alphametics, allowing non-zero integers, whose absolute value does not exceed 11.

Rem 1: The source will be disclosed with the official solution.
Rem 2: An analytical solution (d4) will be applauded!

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Better late than never! Comment 3 of 3 |
Since there are 26  letters in the English alphabet and only Q,Y,Z  are not used in the puzzle -  2*11 numbers  cannot provide a matching set  for 23 unknowns.
Voiding the non-zero restriction and declaring  that O will be set as 0 is therefore a logical rescue assumption.

i therefore  define Charlies's answer as the official solution.

Meglio tardi che mai!

Unfortunately - could not recover the source...
H E L P ,     anyone!

 Posted by Ady TZIDON on 2020-04-15 04:47:13

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