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No. 3


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6 replies to this topic

#1 rookie1ja

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Posted 31 March 2007 - 05:47 PM

No. 3 - Back to the Cryptograms and Algebra Puzzles
Replace the same characters by the same numerals so that the mathematical operations are correct.
RE + MI = FA
DO + SI = MI
LA + SI = SOL

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#2 lyndonjones

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Posted 09 June 2007 - 04:27 PM

Could the answer also be the following ...


1+0 = 1
(RE+MI = FA)

0+0 = 0
(DO+SI = MI)

1+0 = 1
(LA+SI = SOL)

Enjoy the site, came across while solving "Who owns the Fish"!!!
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#3 BoilingOil

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Posted 26 September 2007 - 10:45 PM

Replace the same characters by the same numerals so that the mathematical operations are correct.
RE + MI = FA
DO + SI = MI
LA + SI = SOL



LA + SI = SOL implies S = 1. LA + 1I = 1OL would cause L to be 8 or 9, but in both cases, O would have to be 0, which also is in line with DO + SI = MI
A and I can't both be high enough to get L = 8 and carry over into O, so L must be 9, thus 9A + 1I = 109. A + I must thus be 9
From DO + 1I = MI then follows M = D + 1.

So far we have:
RE + MI = FA
D0 + 1I = MI
9A + 1I = 109

A can not be 0, 1, or 9, which are already in use.
A can not be 8, because I would be 1 (used).
A can not be 7, because then I would be 2, and E 5, but that would leave 3, 4, 6 and 8 which could in no way fit D, F, M and R.
A can not be 6, because both I and E would then be 3.
A can not be 5, because that would make I = 4, and thus E is 1 (used).
A can not be 4, because I would become 5, and thus E would be 9 (used).

Assume A = 3, this would imply I = 6, E = 7. Then 2, 4, 5 and 8 would remain. R = 2, D = 4, M = 5, F = 8 will fit.
Assume A = 2, this would imply I = 7, E = 5. Then 3, 5, 6, and 8 would remain. These can not be put for D, F, M and R in any mathematically correct fashion.

So the correct solution would be S = 1, R = 2, A = 3, D = 4, M = 5, I = 6, E = 7, F = 8, L = 9, O = 0

27 + 56 = 83
40 + 16 = 56
93 + 16 = 109



Thanks again,

BoilingOil
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#4 WonderAlice

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Posted 24 February 2008 - 07:56 PM

I don't know if this solution is correct, but I've found:

FA=9
RE=4
MI=5
DO=3
SI=2
LA=6
SOL=8

So:

RE+MI=FA 4+5=9
SI+DO=MI = 2+3=5
SI+LA=SOL 2+6=8

It's an easy solution. I've found it by deduction, but I'm not sure that I didn't forget to read some condition for the problem (my english isn't very good... to be honest, it's very bad :rolleyes:). If someone has been able to read my strange english, please, say me if I'm right!

Bye!
(It's my first message. I've logged in just now... ^.^)
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#5 rookie1ja

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Posted 24 February 2008 - 08:38 PM

If someone has been able to read my strange english, please, say me if I'm right!

1 letter shall be 1 unique number ... so, for instance, FA can not be just 1 number (9) - it has to be 2 numbers (1 for F and 1 for A)
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#6 aashrai

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Posted 10 December 2010 - 01:56 PM

There is another solution possible assuming that one of the alphabets can be a two digit number (nothing in the question suggests that this is not possible)
O=0
I=2
S=1
L=9
A=7
E=5
M=4
D=3
R=6
F=10
Now we have,
1. RE+MI=FA
65+42=107

2. DO +SI=MI
30+12=42
3. LA + SI=SOL
97+12=109
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#7 aashrai

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Posted 11 December 2010 - 12:14 AM

In the above case, R=8 and F=12 is also a possible solution(the values for the other alphabets are the same)
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