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Cryptarithmetic Problem-2

Ayush Agarwal | 07:55 | 7 comments

Multiplication 

PROB 4

ABC
DEC
FGH
IAC
ACAE
AFAGCH

Solution:

First if you see ( D E 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 ) * 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:

AB2
D62
304
IA2
A2A6
A3A024

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.

152
D62
304
I12
1216
131024

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,

152
862
304
912
1216
131024


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7 comments:

  1. Replies
    1. Which Step..?...pls mention the part i which u faced problems

      Delete
    2. How do we know that carry is 1 in case of c+carry=f?

      Delete
  2. Final result should be 431024 that means HFAGCH, not AFAGCH

    ReplyDelete
  3. MULA please check your answer...!!! It is 131024

    ReplyDelete
    Replies
    1. How do we know that carry is 1 in case of c+carry=f?

      Delete
  4. How do we know that carry is 1 in case of c+carry=f?

    ReplyDelete