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1: When we add two digits its sum is either less than ten or greater than or equal to ten but less than twenty.

if the sum is greater than or equal to ten than the carry over must be one.

in the given example the sum of S and M in thousands column gives the carry over M so M must be 1.

One of the digits in thousands column is M i.e:- 1 . S when added to one has given a carry over. It is possible when S is 9. So S is 9.

S+M=9+1=10

In hundreds column we find that E + O=N

Which is not possible. Hence from tens column there must be a carry over to hundreds column.

E+1=N

it implies that E=N-1 ...........Eqn (1)_

From the tens column N+R+(1)=E+10.............eqn(2)

1 is enclosed with brackets because it is not sure whether there is a carry over from ones or not.

substituting the value of eqn"(1) in eqn (2) we get E + (1)=9

If there is no carry over from ones column than the value of R is 9 but S has already got the value 9 so R cannot be 9 so R=9-1=8

the value of Y cannot be 1 or 0 because the values are already assigned.

so the smallest possible value of y is 2

the numbers left are 3,4,5,6,7

the only possible value of E=5 and N=E+1=5+1=6

E cannot be 7 because if E=7 , N=8 which cannot be as it is already used.

If E=6 than N=7 and D will be 5

if this is true in unit column

D+E=6+5=11

and in the sum the unit digit should be M which is false.

Hence

E=5,N=6 and D=7 such that 7+5=12 and the value of Y in the unit place of the sum is 2.

So

S=0

E=5

N=6

D=7

M=1

O=0

R=8

Y=2

9 5 6 7

+1 0 8 5

1 0 6 5 2

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