CRYPT ARITHMETIC FOR e-LITMUS -2

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PROBLEM NO-3

This type of problems can be solved by back-tracking technique after finding one letter (number).

SOLUTION: 

	M	A	D
		B	E
	M	A	D
R	A	E
A	M	I	D

(M A D) * (E) = (M A D) we can clearly get that E = 1
from the second multiple D * B = E => 1 so only options is 7 * 3 = _1
                                                                                                         3 * 7 = _1
Let’s take B = 3 and D = 7 and E = 1 redraw the table.

	M	A	7
		3	1
	M	A	7
R	A	1
A	M	I	7

M + A = M  then we can understand that the value of A will be 9 because (9 + carry) = 10
Now, A = 9.

	M	A	7
		3	1
	M	A	7
R	A	1
A	M	I	7

See the last column R + (carry) = A [ carry may be equal to 1 because two less than 9 numbers leads to carry (one) ]
So R = 8.          M 9 7
                       *        3 1
                             8  9 1       To get R = 8, M should be equal to 2(M = 2)

	2	9	7
		3	1
	2	9	7
8	9	1
9	2	I	7

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PROBLEM NO-4

SOLUTION :

		S	E	E
			S	O
	E	M	O	O
M	E	S	S
M	I	M	E	O

Clearly if we see the first multiple (S E E) * O = E M O O

From that we can get easily E = 1.

So, replace E with (1) and redraw the table.

		S	1	1
			S	O
	1	M	O	O
M	1	S	S
M	I	M	1	O

If we see the shaded region in above table M + S = M the value of M = 9
(I explained in previous example).

So replace S = 9 and redraw the table.

		9	1	1
			9	O
	1	M	O	O
M	1	9	9
M	I	M	1	O

By the second multiple (9 1 1) * 9 = (M 1 9 9) = (8 1 9 9)

So M = 8 and also see the shaded region O + 9 = _1 so the value of must be equal to 2 (O = 2). Replace the values of M and O and redraw the table.

		9	1	1
			9	2
	1	8	2	2
8	1	9	9
8	3	8	1	2

After replacing M and O value the value of I = 3….