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anonymous
 4 years ago
Fool's cryptarithm of the day,
Here is just another nice problem from an old issue of mathematics exacalibur (
http://www.math.ust.hk/excalibur/v1_n3.pdf
) check out "Qui Trouve Ceci" one.
Good luck!
anonymous
 4 years ago
Fool's cryptarithm of the day, Here is just another nice problem from an old issue of mathematics exacalibur ( http://www.math.ust.hk/excalibur/v1_n3.pdf ) check out "Qui Trouve Ceci" one. Good luck!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0FORTY TEN TEN  SIXTY to get Y + 2N = Y N must be either 0 or 5. if N = 0, no 1 is carried to 2nd column and E must = 5 and 1 is carried to 3rd. if N = 5, 1 is carried to 2nd column and 2E + 1 + T must = T or T + 10. that is 2E = 1 or 2E + 1 = 9 which is not possible so N must be = 0 and E = 5

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1I think ffm was referring to the problem after this one.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1the long division involving Qui Trouve Ceci

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i did manage the above one though  after some sweating!! 29786 850 850  31486

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1The one ffm wanted us to tackle is far more devious (as usual for an ffm problem!)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea  i just looked at it!!!

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1After lots and lots of scribblings and scratching of the head, I have found these sets so far: Q = 1 E = 4, 6 or 8 V = 0, 2 or 4 I = 2, 4, 6 or 8 C = 2, 3, 4, 6, 7 or 8 maybe this will help others progress further but I have hit a brick wall right now. I'll back to this tomorrow with a fresh mind. FFM: Please let me know if I am way off the mark in my findings so far.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Good work jimmyrep :) Here is the solution for that one : http://www.mathematik.unibielefeld.de/~sillke/PUZZLES/ALPHAMETIC/forty+ten+ten=sixty More could be found here: http://www.mathematik.unibielefeld.de/~sillke/PUZZLES/ALPHAMETIC/f @asnaseer: Seems okay to me :)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1Ok, I've got further. These are my current findings: Q = 1 U = 3 I = 2 C = 6 or 7 E = 4 T = 5, 6, 7, 8 or 9 R = 5, 6, 7, 8 or 9 O = 5, 6, 7, 8 or 9 V = 0

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1This was a real messy problem  lots of head scratching, trial and error, and luck, but I think I finally got it: Q = 1 U = 3 I = 2 C = 7 E = 4 T = 9 R = 8 O = 6 V = 0
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