Cryptarithmetic Problem-2

Multiplication 

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

Solution:

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 ofD = (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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