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# Cryptarithmetic Division Problem 2

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

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