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….

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CRYPT ARITHMETIC FOR e-LITMUS

CRYPT ARITHMETIC is a type of mathematical game consisting of a mathematical equation among unknown numbers, whose digits are represented by letters. The goal is to identify the value of each letter. The name can be extended to puzzles that use non-alphabetic symbols instead of letters.

PROBLEM NO-1

			A	B	C
			D	E	C
			F	G	H
		I	A	C
A	C	A	E
A	F	A	G	C	H

SOLUATION

First if you see ( D E C ) * C from this we can confirm that two possibilities:

1)      E = ( 3, 7, 9 ) and C = 5

2)      E = 6 and C = ( 2, 4, 8 )

So now you are in little bit confusion which possibility I should proceed.

If you are not choosing the correct possibility time will be killed off course time is very very precious in eLitmus exam.

So, better step forward to estimate the correct possibility–)

If you see last multiple ( D E C ) * C = E if you assume C = 5 and E = (3, 7, 9 ) the only possible values of D = (2, 4, 8) if you keep any value for D then the value E = 0 that is contradiction.

So it is very clear that first possibilities is wrong then go with the second possibility means

E = 6 and C = (2, 4, 8) Let’s take C = 2 Then the table changes as:

			A	B	2
			D	6	2
			3	0	4
		I	A	2
A	2	A	6
A	3	A	0	2	4

Clearly from the table H = 4 and G = 0 (because G + 2 = 2 means G = 0), and F = 3 (C + carry= F).

After filling this values let’s take (A B 2) * 2 = 3 0 4. It happens only when B = 5 and A = 1

Replace the values and redraw the table.

			1	5	2
			D	6	2
			3	0	4
		I	1	2
1	2	1	6
1	3	1	0	2	4

From the table (1 5 2) * D = 1 2 1 6 It will happen only when D = 8 and from the line (I + 1 + carry = 1) so from that we can say I = 9.

Replace all the values,

			1	5	2
			8	6	2
			3	0	4
		9	1	2
1	2	1	6
1	3	1	0	2	4

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PROBLEM NO-2

		A	S	K
			T	O
	K	A	R	L
O	S	A	K
T	Y	S	O	L

SOLUATION

Let’s see clearly (T O) * K = K from this there will be two possibilities

1)      T = (3, 7, 9) and K = 5

2)      T = 6 and K = (2, 4, 8)

And there is no other clue to estimate which possibility is better to approach first.

Let’s try with the first possibility take T = 3 and K = 5 and draw the table

		A	S	5
			3	O
	5	A	R	L
O	S	A	5
T	Y	S	O	L

Next the only way to approach is 5 * O = L the value for O = (2, 4, 6, 8) if we keep O = 2 then at last

O + carry = T ( 2 + 1 = 3) the value 3 is assigned to two letters it is contradictory.

Let’s take the value of O = 4, T = 7(3 is contradiction) and proceed at last O + Carry = T ( 4 + 1 = 5) the value 5 is assigned to two letters it is also contradictory.

So take value of O = 6 and draw the table.

		A	S	5
			7	6
	5	A	R	L
6	S	A	5
7	Y	S	6	L

From the table it is clear that 5 * 6 = L (5 * 6 = 0) L = 0 and T = 7, R + 5 = 6 so, R = 1.

		A	S	5
			7	6
	5	A	1	0
6	S	A	5
7	Y	S	6	0

(A S 5) * 6 = 5 A 1 0 it will happen only when S = 8 and A = 9 and from the table

5 + S = Y (5 + 8 + carry = 4), Y = 4 let’s replace the table with all values.

		9	8	5
			7	6
	5	9	1	0
6	8	9	5
7	4	8	6	0

THANKS FOR YOUR TIME ........STAY TUNED ....MORE PROBLEM WILL COMING SOON...

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