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How To Begin In Solving A Crypt Arithmatic

Ayush Agarwal | 13:36 | 14 comments

Here We demostrate a small example...............

Let's understand the way how we will solve these questions with an example: 

                F A C
              * H E I
   -----------------
             E E A G
          C J F E 
       D I J D  
 ------------------
        E B F E F G


Step 1:
Determine the characters which can be 0. Although this may not be useful in all cases but keeping the track of it may be useful in many cases.
In our example the multiplicand is FAC and multiplier is HEI
Rule 1: Any leading character will not be 0, so neither of F/H/E/C/D is 0
Rule 2: If any digit among E/I in multiplier is 0 then the corresponding row in multiplication will be all the same digit,
but in our case no row is either EEE/III so neither of E/I is 0
Rule 3: In multiplicand FAC, if C is 0 then the first,second and third row of multiplication will contain the trailing character as C, 
but we don't have trailing character in all row as C, so C is no 0
Result of Rule 1, Rule 2 and Rule 3: Characters which can be 0 are - A/B/G/J
Note: Although the rule says that A can be 0 but if you come across situation where you need to consider one among these suspected character as 0, 
then consider A(middle character of multiplicand) in the last. Or you can assume it to be non zero for simplicity.
Concluding the step 1: None of the character among multiplier, multiplicand and all leading character in each row will be 0. 
So Characters which can be 0 are - B/G/J
Step 2:
Find whether we have any of the following property(in general for all problems):
I * C = I
I * C = C

E * C = E
E * C = C

H * C = H
H * C = C
Remember, they will give problem which has one of these property and this will be starting point of our approach to solve it.
We see that we have the property E * C = E
Now from the rule 3 and 4 we notice that either E is 5 and C is an odd number or E is an even number and C is 6
We will now proceed with the first case when E is 5 and C is an odd number and will consider the second case if we are not able to solve by this assumption. 
Also we will have to think differently from here for each problem but practising more and more problem will make it easier.

              F A C
           * H 5 I
      -------------
          5 5 A G
        C J F 5
     D I J D  
  ---------------
     5 B F 5 F G

Now look carefully, the last column from left says: D + carry = 5. Now think what could be the maximum carry from the last sum C + I = B ?
Remember the sum of two number will never give a carry more than 1 and sum of three numbers will never give a carry more than 2 except in one case:


             U V W
           * X Y Z
      -------------
           9 9 9 9
        9 9 9 9
     9 9 9 9  
    -------------
   1 1 0 9 8 8 9
  
Here the 2nd column from left gives a carry equals to 2 when both numbers of the column are same and are equal to 9(not in our case)
So we see that in D + carry = 5, the carry must be 1 and hence D has to be 4

             F A C
           * H 5 I
     -------------
           5 5 A G
         C J F 5
       4 I J 4  
     -------------
      5 B F 5 F G

Also C can be either 3/7/9 and can't be 1 because if C was 1 then trailing character of each row would have been I,5 and H respectively.
Try to minimize the number of values that a character can have. Let's do it for C. 
Focus on all C in our problem and see how they are generated. Look at second row of multiplication where: 

           F A C
              * 5
          --------
         C J F 5

This multiplication give a number which will always be smaller than 5000 so C can be 3 only.

           F A 3
        * H 5 I
    -------------
        5 5 A G
      3 J F 5
   4 I J 4   
   -------------
   5 B F 5 F G

Focus on the second row of multiplication again: 


         F A 3
           * 5
      --------
      3 J F 5

We can see that F can be either 6 or 7. 
Let's proceed with F = 6 and will consider F = 7 if we are stuck and are not able to proceed with F = 6

             6 A 3
           * H 5 I
     -------------
          5 5 A G
        3 J 6 5
      4 I J 4  
    -------------
     5 B 6 5 6 G

Now see the second column from right A+5=6, which implies that A must be 1 and since we haven't gotten any character yet with value 1, 
accept the value and proceed further.

              6 1 3
           * H 5 I
     -------------
           5 5 1 G
         3 J 6 5
      4 I J 4  
    -------------
     5 B 6 5 6 G 

              6 1 3 
                 * 5
             --------
            3 J 6 5
=> J = 0

             6 1 3
           * H 5 I
      -------------
           5 5 1 G
        3 0 6 5
      4 I 0 4  
     -------------
     5 B 6 5 6 G

              6 1 3 
                 * H
            --------
            4 I 0 4

=> H must be 8 so that when multiplying H with 3 gives unit place digit as 4.


             6 1 3
           * 8 5 I
     -------------
           5 5 1 G
        3 0 6 5
      4 I 0 4  
    -------------
    5 B 6 5 6 G

            6 1 3 
               * 8
          --------
          4 I 0 4

=> I = 9

              6 1 3
           * 8 5 9
       -------------
           5 5 1 G
        3 0 6 5
     4 9 0 4  
     -------------
     5 B 6 5 6 G

From above we can easily find that G = 7 and B = 2

             6 1 3
           * 8 5 9
     -------------
          5 5 1 7
       3 0 6 5
    4 9 0 4  
   ------------
    5 2 6 5 6 7




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14 comments:

  1. in step 2 ,after concluding C as 3,Y F has to be 6 or 7???

    ReplyDelete
    Replies
    1. because FA3*5 is 3JF5 that means the leading character is 3 if we multiply 5 with 6 we get 30 and 5 with 7 we get 35. so, the leading character can be 3 with only two possibilities that is 6 or 7.

      Delete
  2. Good..........PLEASE upload some more Crypts ques.

    ReplyDelete
  3. "FAC * 5 = CJF5
    This multiplication give a number which will always be smaller than 5000 so C can be 3 only."

    Sir, how can u say this????

    ReplyDelete
    Replies
    1. FAC*5
      FACis three digit no.so it must be less than 1000.
      Now 1000*5=5000. So three digit num*5 less 5000.
      1 st confusion solved.
      Now C*5=5 thats mean C is odd num.
      CJF5 ,C is not 5 as some other digit already is 5.
      And CJF5 is less than 5000.so C is 3 only (C=1 not possible. See others conditions.)

      For any further confusion comment below.

      Thanks,
      AllAboutElitmus Team

      Delete
  4. thanx for such a good explanation

    ReplyDelete
  5. i never have an idea about this crypt problems
    im facing difficulty please suggest me what i should do to crack this section

    ReplyDelete
  6. sir how to solve this kind of ques..
    LQL*PVS
    ----------
    L?QV
    K3P?
    BL2K
    --------
    EUKQSV

    ReplyDelete
  7. This comment has been removed by the author.

    ReplyDelete
  8. in step 2 ,after concluding C as 3,why F has to be 6 or 7???

    ReplyDelete
  9. We know F can be 0/1/2.../9 but F cannot be 3/4/5 becoz they have been used.Now F cannot be 0 as it was proved earlier,now F cannot be 1/2 becoz max. value obtained cannot be greater than 3000.Now similarly F cannot be 8/9.So we are left with only two values which are 6/7.

    ReplyDelete
  10. i dnt hv idea of ths cryparthmatic plz exam agn plzz

    ReplyDelete