Answer Posted / tamilarasan

consider all the chars are unit and distinct digit
then E can only hold 3,6 or 9
so n can hold only 1,2 or 3 respectively

like wise 'O' can hold only 7,4 or 1 respectively

if u take as E=3, N=1, O=7 then U also take the value as 3, it overcomes the rule
likewise
if u take as E=9, N=3, O=1 then U also take the value as 3, it also overcomes the rule

so take E=6,N=2,O=4
then

2 4 4 2
S 4 4 2
M 4 4 2
----------
J 3 2 6
----------
so U=3
now only 1,5,7,8,9 are remaining

in the thousands digit
we have

1+2+S+M=J 1 is carry while adding 4+4+4
so j can have at the maximum of 9
and s+M can have the maximum of 6
so S and M can have only 1 and 5
so S=1,M=5,and J=9

so
2 4 4 2
1 4 4 2
5 4 4 2
----------
9 3 2 6
----------

HERE 0 is excluded

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