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Each letter is a distinct single digit number.

It is given that M ≠ 0

Solve for every letter, also solution is unique.

It is given that M ≠ 0

Solve for every letter, also solution is unique.

Answer:

9 5 6 7

+ 1 0 8 5

---------

1 0 6 5 2

Explanation:

Maximum value of 2 single digit is 18.

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)

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)

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 )

Now E = 5 or 6

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

Y + 10 = D + E

E + 1 = N

And finally, we have,

D = 7,

E = 5,

M = 1,

N = 6,

O = 0,

R = 8,

S = 9,

Y = 2

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