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rookie1ja

No. 3

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rookie1ja    11

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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No. 3 - solution

27 + 56 = 83

40 + 16 = 56

93 + 16 = 109

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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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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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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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rookie1ja    11
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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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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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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