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I got the same answers, this is how i did it.

 

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M+5+E= *5

A+N+V+*=()E

5+A+E+()="V

A+"=M0

 

let's assume that all letters are integers from 0-9

 

let's also assume you can't use the same letter twice.

 

therefore A+" must = 10 therefore M=1

 

1+5+E= *5 -> 6+E=*5 * must be one and E must be 9 if statement 1 is true

 

A+N+V+1=()9

 

ANV must be 378 not necessarily in that order. (only numbers left that can give that sum if statement 1 and 2 were correct.) ()= 1

 

5+A+9+1="V V=[3,7 or 8]

 

15+A = "[3,7 or 8] A=8 " = 2 V=3

 

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The way I did it:

[hide]M must be 1, as it is an extra place out. Then M + D + E = D, so M + E is 0, so E is 9. Then, A must be 8, as that is the only one that will give the 1 for M. D + A + E + carry(A + N + V) = V + 20. A + N + V + 1 = E + 10*carry, so 9 + N + V = 9 + 10*carry, so N + V = 10, and the carry is 1. 9 + 8 + 5 + 1 = V + 20, so V = 3, and N = 7.

O = 0

M = 1

V = 3

D = 5

G = 6

N = 7

A = 8

E = 9

MOVED = 10395.

[/hide]

=Uncool-

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