Cryptarithmetic Division Problem 3

PT

As,
 A N M L
 x     D
--------
 A N M L
thus, D = 1.
         C 1 N K B 
         _______________
A N M L |P K N L B C 1 E
         P 1 B A
         --------------
           L C B B
           A N M L
           --------------
           C M M M C
           C C B K N
           -----------------
             B K C L 1
             B M N B M
             -----------------
               1 L B N E
               1 B L E 1
               -----------------
                 B M A P
Check Unit digit method here.

Now, Find the multiplication with most substituted values,
it is -
  A N M L
  x     B
--------
1 B L E 1

Now, using Unit digit's method possible ways of getting 1 are
(check the link above for Unit digit method to find all rules)

3 x 7 = _1 (21 only last digit considered)
7 x 3 = _1 (21 only last digit considered)

Thus, values are 
either L = 3, B = 7 and L = 7, B = 3

using case 1 first, 

  A N M 3
  x     7
---------
1 7 3 E 1

Now, if you consider A x 7, this results in 17.
There is only one possibility, for A x 7 + carry = 17

when A = 2 and carry = 3( A = 1 not possible as
7 + 6(max carry ie 7 x 9 =63) results only in 13 which is <17.

So now eqn is 

  2 N M 3
  x     7
---------
1 7 3 E 1

Now, we know what carry from 7 x N was 3, this is only possible when

7 x N = _3 when N = 4( 28 + some carry >=2) 
or 5( 35 + some carry less equal to 4)

Taking N = 4,

  2 4 M 3
  x     7
---------
1 7 3 E 1

Now, at unit's digit 3 value and 3 as carry.
It is possible for 7 x 4 + 5(carry) = 33

Now to get 5 carry from last step 
you have to multiply 7 with either 7 or 8

Now, take M = 8

  2 4 8 3
 x      7
---------
1 7 3 E 1

Now carry from unit's 7 x 3 will be 2
and 7 x 8 + 2 carry = give value of E = 8. and M also 8. So, not possible.
Sim. M = 7 is not possible(try yourself)

Now try L = 7 and B = 3

  A N M 7
  x     3
---------
1 3 7 E 1

Now A = 4 to get 13 with 1 carry from prev. multiplication,
  1
  4 N M 7
  x     3
---------
1 3 7 E 1

Now, to give 1 carry and 7 at unit's place
value of N = 5 with 2 carry from previous operation

  1 2
  4 5 M 7
  x     3
---------
1 3 7 E 1

Now, 3 x M can give 2 carry for M={6, 8, 9}
Let's do hit and trial. Take M = 6.

  1 2 2
  4 5 6 7
  x     3
---------
1 3 7 E 1

Now, 3 x 7 gives 2 carry to next mult.
and 

3 x 6 + 2 as carry = 20 so E = 0.

  1 2 2
  4 5 6 7
  x     3
---------
1 3 7 0 1

E = 0, D = 1, A = 4, N = 5, M = 6 its easy to find other values too. 

PT