# 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 HNow, converting into Mult. problem -   E F G  x   E--------    - 11 H 0 Eand Now, G x E = _EBy hack 3 or Rule 3 check this pageWe get two cases -Case 1 : G = {3, 7, 9} E = 5or G = 6 and E = {2, 4, 8}Taking Case 1.1 E = 5 and G = 3 -   E F 3  x   5--------1 H 0 5Now, this is rejected as 3 x 5 = 15, 1 carryand for any number P x 5 + 1(carry) can't give 0 at units placei.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 5Now, 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 0Taking case 2.1 E = 2 and G = 6  2 F 6  x   2--------1 H 0 2Now, this again is rejected as 6 x 2 gives 1 carryand 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 4Now, 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 4now, clearly 7 x 4 + 2 = 30 gives 3 carry4 x 4 + 3 = 19 thus H = 9 and easily you can find other values as well.`

PT