For sure E=1 as MAD*E=MAD,
M+A=M carry of 1will be generated so A=9 and hence R=8
And so you can try for value of B and M that will give R as 8,so M=2 and B=3
Then I=0 as A+E=10
Finally trying for D wll get its value as 7.
297
*31
=9207

297*31
-----------
297
891
----------
9207

297*31
-----------
297
891
----------
9207

297
31
---------
297
891
-------
9207

297
*31
-----------
297
891
---------
9207
-----------
is the correct answer....

btw the question is not written in order...

M A D
* B E (Its a multiplication * wont be in question)
---------
M A D
R A E 0
------------
A M I D

E=1*MAD=MAD 1st step

B*D=(sth)1 cuz B*D=(sth) E and E=1 so B/D are 3/7 (dont know which one is which in this step)

now possibilities

if(B==3){
if(A==4){
B*A=3*4=12+2=14 // (+2 cuz carry is 2 since 3*7=21) (B*A must give (sth)A in this case 14=(sth)4)
}
if(A==9){
B*A=3*9=27+2=29
}
}
if(B==7){
B*A=7*8=56+2=58
}

So...A=4||A=9||A=8


3rd step in addition M+A=M
so...we conclude there is a carry in this step...
cuz suppost M=3 and A=4 so... M+A can never give M so the number M+A must be greater than 10 Eg. is M=3 from addition we should get 13

suppose A=9 so B=3 and D=7

we know E=1
carry 1 from previous addition
so possibilities for M are
if M==1
9+1+1=11

if M==2
9+1+2=12

if M==3
9+1+3=13
...
...till M=9
(last digit has to be M)
but 0/1/3/7/9 are already known so possibilities for M are 2/4/5/6/8 (I=0 since A+E=9+1=10)

Now last step in addition R+carry=A
so possibility for carry in this step is just one i.e 1 cuz from last step M+A gives max 9+9=18(just example)
so R+1=9
so R=8

remaining digits 2/4/5/6

Now last Multiplication

B*M=R i.e 8
we know B=3
so
3*2=6+2(carry from previous step)=8 i.e R
so M=2

Now M=2, A=9, D=7, B=3, E=1

2 9 7
* 3 1
----------
2 9 7
8 9 1 X
---------
9 2 0 7

297*31=9207

:) seriously this is no where close to the level of elitmus