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# No. 3

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

This old topic is locked since solution is already provided in the Spoiler below.
Pls visit New Puzzles section to see always fresh brain teasers.

Spoiler for Solution

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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 ). 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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