Can you break this code?

$\begin{array} { l l l l l } & S & E & N & D \\ + & M & O & R & E \\ \hline M & O & N & E & Y \\ \end{array}$

How $O=0$ we have:

$\begin{array} { l l l l l } & S & E & N & D \\ + & M & 0 & R & E \\ \hline M & 0 & N & E & Y \\ \end{array}$

We can conclude that $M=1$ and $S=9$ because if $S<9$ in the Thousands column need a carry up what is not possible. We have now:

$\begin{array} { l l l l l } & 9 & E & N & D \\ + & 1 & 0 & R & E \\ \hline 1 & 0 & N & E & Y \\ \end{array}$

Because $N\neq E$ , the Hundreds column need a carry up. If $N=E+1$ , $R$ Should be $9$ . It's not possible. Thus $9>R>1$ and the Tens column needs a carry up. We have now $3$ columns with the number $E$ .

$N$ can't be $8$ because if so, $E=9$ . Thus $8>N>1$ .

If the Tens column have a carry up, $D+E=10+Y$ . And the Hundreds column needs a carry up too. Thus $1+N+R=E+10$ or $N+R=E+9$ .

Using the last equations we can conclude that $E+1+R=E+9$ or $E+R=E+8$ or $R=8$ . We found another number! Let's put it in the sum:

$\begin{array} { l l l l l } & 9 & E & N & D \\ + & 1 & 0 & 8 & E \\ \hline 1 & 0 & N & E & Y \\ \end{array}$

If $N=E+1$ and $D+E=10+Y$ we have: $D+N-1=10+Y$ or $D+N=Y+9$ .

How $Y\neq0$ and $Y\neq1$ , $Y>1$ . The numbers $8$ and $9$ were found in the last operations, $Y$ can be only in the interval $8>Y>1$ .

Observe that, to $Y=7$ , the only values for $D$ and $E$ are $8$ and $9$ .

To $Y=6$ , we have the same problem because $E<8$ , and the only values for $D$ and $E$ is too large and no match the rules.

The same problem happens with the numbers from $3$ to $5$ . Because if you try to put this numbers in the others columns the results will be a contradiction like $Y=4$ . The Only numbers that match it are for $D$ and $E$ are $6$ and $8$ , $5$ and $9$ . These numbers are not a possible solution.

Thus the value of $Y$ should be $2$ :

$\begin{array} { l l l l l } & 9 & E & N & D \\ + & 1 & 0 & 8 & E \\ \hline 1 & 0 & N & E & 2 \\ \end{array}$

The only values for $D$ and $E$ if $Y=2$ are $7$ and $5$ respectively. It can be verified using the equations more above.

Now:

$\begin{array} { l l l l l } & 9 & 5 & N & 7 \\ + & 1 & 0 & 8 & 5 \\ \hline 1 & 0 & N & 5 & 2 \\ \end{array}$

Finally, if $N=E+1$ , $N=6$ .

The last sum with the result is the following:

$\begin{array} { l l l l l } & 9 & 5 & 6 & 7 \\ + & 1 & 0 & 8 & 5 \\ \hline 1 & 0 & 6 & 5 & 2 \\ \end{array}$

I know my solution is very long, and I used trial and error and some parts, but it is the way that I solved it. If have a easier method, can you explain putting your solution to this problem too?

Note: I have assumed that we don't know that O = 0, it's pretty straightforward to arrive at that. I suppose that piece of information (O = 0) can be removed from the question.

The maximum possible first digit of the 5 digit number formed by adding two 4 digit numbers is 1 i.e. 9999 + 9999 = 19998

M = 1

There has to be a carryover over the ten thousand's place. Since M = 1, S = 8 or 9. And correspondingly, O = 0 or 1. O cannot be 1 since M already = 1.

O = 0

Over the hundred's place, we cannot have E + 0 = N since E and N are distinct digits. There should be a carryover i.e. E + 1 + 0 = N.

There will not be a carryover over the thousand's place except in the case of E = 9. But if that happens, N = 0 when O already = 0. So E is not = 9 and there is no carryover over the thousand's place. Which eliminates the possibility of S = 8.

S = 9

As we observed previously, N = E + 1. Consider the ten's place.

If there's no carry over, E + 1 + R = 10 + E, giving R = 9 which is not possible as S already = 9. If there's a carry over, 1 + E + 1 + R = 10 + E, which gives R = 8.

R = 8

Now the remaining digits to be used are 2, 3, 4, 5, 6, 7. For a carryover to be present over the ten's place, Y = 2 (not 1 or 0, they've already been assigned)

Y = 2

For Y = 2, there is only 1 possible combination of D and E i.e. 5 and 7. But if E = 7, then N= 8 which is not possible as R already = 8.

E = 5

D = 7

N = 6

That gives us our answer MONEY is 10652

S+M = 10, therefore M = 1 and S = 9.

N = E + 1

If N + R = E + 10 => E + 1 + R = E + 10 and R = 9 which is impossible, therefore:

N + R + 1 = E + 10 => E + 2 + R = E + 10 => R = 8

D, E and N are all distinct, so for D + E to exceed 10, one is 5, one is 6, and one is 7. E is 5 or 6 and N is 6 or 7 - if E = 6, N = 7 and D = 5, which would mean Y = 1, which is not possible, so E = 5, N = 6 and D = 7, and Y = 2. The sum then becomes:

9567 + 1085 = 10652

10000M+100N+10E+Y=1000S+100E+10N+D+1000M+10R+E

9000M+90N+Y=1000S+D+10R+91E

M must be 1. The thousand column is 0 in the end, and the hundred column does not carry-over.So S must be 9.

so we get: 90N+Y=D+10R+91E

10(9N-R-9E)=D-Y+E

The right part of the equation must be smaller than 20 (since 9+9-1=17), so 9N-R-9E=1, D-Y+E=10

So 9N-9E=R+1,R+1 can be divided by 9,so R=8, N-E=1

And now we start to try,because all numbers are distinct, only 2,3,4,5,6,7 are available.

When we try N=6, E=5,D=7,Y=2 ,it fit the equation above.(N-E=1,D-Y+E=10)

So MONEY=10652

I applied following logic:

1. Sum of two positive numbers can generate a carry of 1, so M=1 .

2. S + M should be 10, also given that

      S     E
      M     0
    -----------
      0     N
   

Since E on addition with 0 is giving a different result which is not 0 therefore that result cannot generate a carry forward for next sum (S+M), therefore S=9 .

1. E + 0 is giving N that can only mean that E and N are consecutive numbers. N + R is giving E plus 1 as carry forward to next digit's place, that is only possible when N is added with 9, now R cannot be 9 since the numbers are "distinct", hence R =8 .

2. Now we have three numbers left D E and N, we know that E and N are consecutive also sum of D and E should generate a carry forward. The only combination that satisfies the above conditions and condition for all numbers to distinct is E=4, N=5 and D=7 .