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Cryptarithmetic Division Problem 2

Cryparithmetic Division Problem 4

Lets go ahead in this problem by converting into multiple multiplication problems – 

But before that it is quiet evident that K = 1 as EFG is giving EFG in row 2 and row 4. 

Also in row 1 and row 2 G – G gives A, thus A = 0,  so the problem becomes –

PT

       1 1 E F
______________
E F G |D E G B C 1
E F G
--------------
F 0 B
E F G
--------------
C C F C
1 H 0 E
-----------------
B G I 1
B B B C
-----------------
B E H

Now, converting into Mult. problem -

E F G
x E
-------- - 1
1 H 0 E

and

Now, G x E = _E

By hack 3 or Rule 3 check this page

We get two cases -
Case 1 : G = {3, 7, 9} E = 5
or G = 6 and E = {2, 4, 8}

Taking Case 1.1 E = 5 and G = 3 -


E F 3
x 5
--------
1 H 0 5

Now, this is rejected as 3 x 5 = 15, 1 carry
and for any number P x 5 + 1(carry) can't give 0 at units place
i.e in our question 5 x F + 1 cant give 0 at units place,


Take Case 1.2 E = 5 and G = 7 :

E F 7
x 5
--------
1 H 0 5

Now, 5 x 7 = 35 and 3 carry,
Now, 5 x F + 3 can't give 0 again, thus rejected.


Trying case 1.3 E = 5 and G = 9:

Now, this is also rejected as 4 will be carry and 5 x F + 4 can't give 0

Taking case 2.1 E = 2 and G = 6

2 F 6
x 2
--------
1 H 0 2

Now, this again is rejected as 6 x 2 gives 1 carry
and F x 2 + 1 can't give 0. The number will always be odd.

Trying case 2.2 E = 4 and G = 6:

4 F 6
x 4
--------
1 H 0 4

Now, 4 x 6 = 24 gives 2 carry,
and F x 4 + 2 can give 0 when F = 7

replacing
4 7 6
x 4
--------
1 H 0 4

now, clearly 7 x 4 + 2 = 30 gives 3 carry
4 x 4 + 3 = 19 thus H = 9

and easily you can find other values as well.

PT