# Send More Money Each letter is a distinct single digit number.

It is given that M ≠ 0

Solve for every letter, also solution is unique.

9 5 6 7
+  1 0 8 5
---------
1 0 6 5 2

Explanation:
Every letter is distinct

Maximum value of 2 single digit is 18.

hence M = 1

there are 2 possibilities
case 1: there is no carry
case 2: carry is 1 (carry cannot be more than 1 as maximum 2 single digit sum is 18)

In column 4, let us solve for 'O' and 'S'
Assume Case 1
S + M = 10 + O
S + 1 = 10 + O
S = 9 + O

Assume Case 2
carry + S + M = 10 + O
1 + S + 1 = 10 + O
S = 8 + O

Here S is single digit (given)
Therefore maximum value of S could be 9.
So, we have 2 case: O = 0 or O = 1

But we have M = 1 (given that every letter is distinct)
so O = 0
and S = 9
And we can conclude that, there is not any carry ..............(1)

Look at column 3

Assume Case 1
E + O = N
E + 0 = N
E = N (which is not possible, as every letter is distinct)

Assume Case 2
carry + E + O = N
1 + E = N .................(2)
E < 9 because if E = 9 then E + 1 = 10 but column 3 cannot produce any carry as per result (1)

From this we know there is carry from column 2.

Look at column 2

As, there is carry from column 2 E > 9

Assume Case 1
N + R = 10 + E
1 + E + R = 10 + E (from (2) we know N = 1 + E)
so R = 9 (which is not possible because S = 9)

Assume Case 2
carry + N + R = 10 + E
0 + 1 + E + R = 10 + E (from (2) we know N = 1 + E)
R = 8

From this we know there is carry from column 1.

Look at column 1

Column 2 has a carry so Y > 9
but Y ≠ 10 or 11 (because O = 0, M = 1) so Y > 11
D,E ≠ 8 or 9 (because R = 8, S = 9)

So, D,E = 7,5,6
7 + 5 = 12
7 + 6 = 13
5 + 7 = 12
6 + 7 = 13

If E = 7 then 1 + E = N and N = 8 (which is not possible because R = 8 )

Therfore, D = 7

Now E = 5 or 6
If E = 6 then 1 + E = N and N = 7 (which is not possible because D = 7 )

Therfore, E = 5

Y + 10 = D + E
Therfore, y = 2

E + 1 = N
Therfore, N = 6

And finally, we have,

D = 7,
E = 5,
M = 1,
N = 6,
O = 0,
R = 8,
S = 9,
Y = 2