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

- schrodinger

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- anonymous

FORTY
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

I think ffm was referring to the problem after this one.

- asnaseer

the long division involving Qui Trouve Ceci

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## More answers

- anonymous

oh yes

- anonymous

i did manage the above one though - after some sweating!!
29786
850
850
-----
31486

- asnaseer

The one ffm wanted us to tackle is far more devious (as usual for an ffm problem!)

- anonymous

yea - i just looked at it!!!

- asnaseer

After 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

Good work jimmyrep :) Here is the solution for that one :http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/ALPHAMETIC/forty+ten+ten=sixty
More could be found here:http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/ALPHAMETIC/f
@asnaseer: Seems okay to me :)

- asnaseer

Ok, 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

This 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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