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Solving elitmus Cryptarithmetic Questions in Logical Reasioning Section-Method-I


In  elitmus test you will be getting 3 questions(30 marks) on cryptic multiplication...

This tutorial will be very helpful in solving those questions.I myself solved those 3 questions just by reading this tutorial. you won't find any help regarding this topic anywhere else so please read this tutorial.   




HOW TO SOLVE A PUZZLE

1. Preparation

Rewrite the problem, expanding the interlinear space to make room for
trial numbers that will be written under the letters.

For example, the puzzle SEND + MORE = MONEY, after solving, will appear like
this:

S
E
N
D

9
5
6
7





+
M
O
R
E

1
0
8
5
---------------
M
O
N
E
Y
1
0
6
5
2


2. Remember cryptarithmetic conventions
  • Each letter or symbol represents only one digit throughout the
    problem;
  • When letters are replaced by their digits, the resultant arithmetical
    operation must be correct;
  • The numerical base, unless specifically stated, is 10;
  • Numbers must not begin with a zero;
  • There must be only one solution to the problem.
3. See subtractions as "upside-down" additions

Ease the analysis of subtractions by reading them as upside-down additions. Remember that you can check a subtraction by adding the difference and the subtracter to get the subtrahend: it's the same thing. This subtraction:
          C O U N T
           - C O I N
           ---------
              S N U B
must be read from the bottom to the top and from the right to the left, as if it were this series of additions:
        B + N  = T + C1
        U + I   = N + C2
        N + O = U + C3
        S + C  = O + C4
C1, C2, C3 and C4 are the carry-overs of "0" or "1" that are to be added to the next column to the left.

4. Search for "0" and "9" in additions or subtractions

A good hint to find zero or 9 is to look for columns containing two or three identical letters.
Look at these additions:
      * * * A            * * * B
  + * * * A         + * * * A
    -------           -------
    * * * A           * * * B
The columns A+A=A and B+A=B indicate that A=zero. In math this is called the "additive identity property of zero"; it says that you add "0" to anything and it doesn't change, therefore it stays the same. Now look at those same additions in the body of the cryptarithm:
     * A * *           * B * *
  + * A * *         + * A * *
    -------           -------
    * A * *           * B * *
In these cases, we may have A=zero or A=9. It depends whether or not "carry 1" is received from the previous column. In other words, the "9" mimics zero every time it gets a carry-over of "1".

5. Search for "1" in additions or subtractions

Look for left hand digits. If single, they are probably "1".
Take the world's most famous cryptarithm:
             S E N D
         + M O R E
         ---------
         M O N E Y
"M" can only equal 1, because it is the "carry 1" from the column S+M=O (+10). In other words, every time an addition of "n" digits gives a total of "n+1" digits, the left hand digit of the total must be "1". In this Madachy's subtraction problem, "C" stands for the digit "1":
        C O U N T
         - C O I N
         ---------
           S N U B

6. Search for "1" in multiplications or divisions

In this multiplication:
         M A D
             B E
         -------
           M A D
         R A E
         -------
         A M I D
The first partial product is E x MAD = MAD. Hence "E" must equal "1". In math jargon this is called the "identity" property of "1" in multiplication; you multiply anything by "1" and it doesn't change, therefore it remains the same.Look this division:
                 K T
            --------
  N E T /    L I N K
                 N E T
             ----------
                   K E K K
                   K T E C
             ----------
                       K E Y
In the first subtraction, we see K x NET = NET. Then K=1.

7. Search for "1" and "6" in multiplications or divisions

Any number multiplied by "1" is the number itself. Also, any even number multiplied by "6" is the number itself:
         4 x 1 = 4
         7 x 1 = 7
         2 x 6 = 2 (+10)
         8 x 6 = 8 (+40)        
Looking at right hand digits of multiplications and divisions, can help you spot digits "1" and "6". Those findings will show like these ones:
                                      C B
                                      -------
     * * A                * * A / * * * * *
       B C                            * * * C
 --------                        ------
   * * * C                            * * * *
 * * * B                              * * * B
 ---------                      -------
 * * * * *                               * * *
The logic is: if
   C  x  * * A  =  * * * C
   B  x  * * A  =  * * * B
then A=1 or A=6.

8. Search for "0" and "5" in multiplications or divisions

Any number multiplied by zero is zero. Also, any odd number multiplied by "5" is "5":
          3 x 0 = 0
          6 x 0 = 0
          7 x 5 = 5 (+30)
          9 x 5 = 5 (+40)        
Looking at right hand digits of multiplications and divisions, can help you spot digits "0" and "5". Those findings will show like these ones:
                                            C B
                                         ----------
    * * A                   * * A / * * * * *
      B C                                 * * * A
  -------                          ---------
  * * * A                                 * * * *
* * * A                                    * * * A
---------                            -------
* * * * *                                      * * *
The logic is: if
   C  x  * * A  =  * * * A
   B  x  * * A  =  * * * A
then A=0 or A=5

9. Match to make progress

Matching is the process of assigning potential values to a variable and testing whether they match the current state of the problem.
To see how this works, let's attack this long-hand division:
                      K M
               ----------
    A K A / D A D D Y
                D Y N A
               ---------
                 A R M Y
                 A R K A
              -------
                       R A
  
To facilitate the analysis, let's break it down to its basic components, i.e., 2 multiplications and 2 subtractions:
   I.          K  x  A K A = D Y N A
 
  II.          M  x  A K A = A R K A
  
  III.         D A D D
            - D Y N A
            ---------
                A R M 

  IV.         A R M Y
            - A R K A
            ---------
                  R A
From I and II we get:
   K  x  * * A  =  * * * A
   M  x  * * A  =  * * * A
This pattern suggests A=0 or A=5. But a look at the divisor "A K A" reveals that A=0 is impossible, because leading letters cannot be zero. Hence A=5.Replacing all A's with "5", subtraction IV becomes:
           5 R M Y
        - 5 R K 5
        ---------
                  R 5
From column Y-5=5 we get Y=0. Replacing all Y's with zero, multiplication I will be:
K  x  5 K 5  = D 0 N 5
Now, matching can help us make some progress. Digits 1, 2, 3, 4, 6, 7, 8 and 9 are still unidentified. Let's assign all these values to the variable K, one by one, and check which of them matches the above pattern. Tabulating all data, we would come to:
                        K  x  5K5  =  D0N5
                ----------------------
                        1     515      515
                        2     525     1050
                        3     535     1605
                        4     545     2180
                        6     565     3390
SOLUTION --> 7     575     4025   <-- SOLUTION
                        8     585     4680
                        9     595     5355
               ----------------------
      
You can see that K=7 is the only viable solution that matches the current pattern of multiplication I, yielding:
    K  x  A K A  =  D Y N A
   7     5 7 5     4 0 2 5
This solution also identifies two other variables: D=4 and N=2.


10. When stuck, generate-and-test

Usually we start solving a cryptarithm by searching for 0, 1, and 9. Then if we are dealing with an easy problem there is enough material to proceed decoding the other digits until a solution is found.
This is the exception and not the rule. Most frequently after decoding 1 or 2 letters (and sometimes none) you get stuck. To make progress we must apply the generate-and-test method, which consists of the following procedures:
  • 1. List all digits still unidentified;
  • 2. Select a base variable (letter) to start generation;
  • 3. Do a cycle of generation and testing: from the list of still
    unidentified digits (procedure 1) get one and assign it to
    the base variable; eliminate it from the list; proceed guessing values for
    the other variables; test consistency; if not consistent, go to perform the
    next cycle (procedure 3); if consistent, stop: you have found the solution
    to the problem.
To demonstrate how this method works, let's tackle this J. A. H. Hunter's addition:
           T A K E
                  A
       +  C A K E
       ----------
          K A T E
The column AAA suggests A=0 or A=9. But column EAEE indicates that A+E=10, hence the only acceptable value for "A" is 9, with E=1. Replacing all "A's" with 9 and all "E's" with 1, we get
           T 9 K 1
                9
       +  C 9 K 1
       ----------
          K 9 T 1
Letter repetition in columns KKT and TCK allows us to set up the following algebraic system of equations:
C1 + K + K = T + 10
       C3 + T + C = K        
Obviously C1=1 and C3=1. Solving the equation system we get K+C=8: not much, but we discovered a relationship between the values of "K" and "C" that will help us later. But now we are stuck! It's time to use the "generate-and-test" method. Procedure 1: digits 2,3,4,5,6,7 and 8 are still unidentified;
Procedure 2: we select "K" as the base variable;

CYCLE #1, procedure 3: column TCK shows that T+C=K and no carry, hence "K" must be a high valued digit. So we enter the list obtained through procedure 1 from the high side, assigning "8" to the base variable "K".
Knowing that K+C=8, if K=8 then C=0. But this is an unacceptable value for "C", because the addend "CAKE" would become "0981" and cryptarithmetic conventions say that no number can start with zero. So, we must close this cycle and begin cycle #2.
By now, the addition layout and the table summarizing current variable data would look like this:
T 9 8 1   CYCLE  A   E   K   C   T
         9   ========================
+  0 9 8 1    #1    9   1   8  [0]
----------
   8 9 T 1
Conflicting values for variables are noted within square brackets.

CYCLE #2, procedure 3: assigning "7" to the letter "K" we get C=1 because K+C=8. This is an unacceptable value for "C" considering that we have already fixed E=1. Again we have to close the current cycle and go to cycle #3, with the setup and table showing:
     T 9 7 1      CYCLE  A   E   K   C   T
         9                    ========================
+  1 9 7 1        #1       9   1   8  [0]
----------     #2       9   1   7  [1]
   7 9 T 1

CYCLE #3, procedure 3: assigning "6" to the letter "K" we get C=2 because K+C=8. Testing these values for "K" and "C" in the column TCK, we get C3+T+2+=6 making T=3.Now, testing T in column KKT, we would obtain C1+K+K=T+10 or 1+6+6=T+10, making T=3. This is an acceptable value for T, confirming the previous value T=3 we had already found.
So, we have got the final solution to the problem, stopping the routine "generate-and-test".
The final layout and table would read

     3 9 6 1               CYCLE      A      E      K            C      T
         9                  ========================
+  2 9 6 1                  #1          9             8 [0]
----------               #2          9       1        7[1]
   6 9 3 1                   #3          9        1       6     2      3   


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+ comments + 54 comments

19 December 2013 at 05:22

I am so confused.... please tel me how u allocate the numbers for each letters....

22 December 2013 at 20:56

I am so confused ............ please tell me how u allocate numbers to each letter

22 December 2013 at 20:56

please reply me soon caz i hav elitmus on jan 5th

13 January 2014 at 10:08

@bhuvana call me on this no.7533048011,
I have learn you how to solve this puzzle.

21 January 2014 at 10:24

thnks!!!!!!

3 February 2014 at 10:00

hi this is very helpfull thanku :-)

8 February 2014 at 05:31

its really very helpful,,thankss

8 February 2014 at 12:55

how can it be solved..much confusing

11 February 2014 at 09:05

[-(

20 February 2014 at 22:51


PLS SOLVE

N Y N N E
+ L O O K S
= S L E E P Y

7 March 2014 at 02:41

eyo...so much confusing it is....eeeeeeeeeeeeeeeeeeeeeee

7 March 2014 at 02:43

;((

10 March 2014 at 22:40

how are we supposed to allocate the digits to the alphabets...what is the procedure for such decoding...plzzz telllll...!!!!! :-?

11 March 2014 at 10:37

helpfull... but confusing... :-b

13 March 2014 at 09:42

plz tel me how 2 solve for
CROSS+ROADS=DANGER;
TWO+TWO=FOUR

16 March 2014 at 06:14

[-( :-? :-s

16 March 2014 at 06:16

Thanks!!! It was very useful! (h)

18 March 2014 at 18:36

So confusing.......👎👎👎👎👎👎👎👎👎

22 March 2014 at 10:18

hey Ruhi can u help me?? actually i m trying many times but i don't understand plz help me

22 March 2014 at 10:20

hey Ankita can u help me?? actually i m trying many times but i don't understand plz hel me

22 March 2014 at 10:30

Reply me soon because my exam on head please..

26 March 2014 at 08:23

Need Help.... If WORLD + TRADE = CENTER THEN C+E+N+T+E+R=?
(HOW TO ALLOCATE DIGITS TO ALPHABETS?)

26 March 2014 at 13:24

(h)

7 April 2014 at 07:53

(h)
These is very helpful tq.

7 April 2014 at 08:02

but totaly confusing ;((

9 April 2014 at 00:48

(h) thanks sir ..its really very helpful....... :)

12 April 2014 at 04:26

good job 8-)

21 May 2014 at 02:50

don get it any1 can help me?? plzz i don unserstand these things....:(

5 June 2014 at 06:09

not possible

19 June 2014 at 23:09

i love cryptarithmetic (k)

20 June 2014 at 13:35

Girish plz send me study meterial regrading elitmus . rupeshkumargdr@gmail.com and even cryptari... thanks

4 July 2014 at 02:00

Two+two=four
734+734=1468

11 July 2014 at 10:36

Send me study material for elitmus mail I'd is rakeshgoldenstar11@gmail.com Girish send the material pls

13 July 2014 at 01:46

Sir can you please send me the cryptarithmetic material on ankitkulshreshtha12@gmail.com. I have to prepare for e litmus

24 July 2014 at 21:46

The eventual intend of the management Cat Entrance Exam candidates in to have a high achieve with good location and percentile, in order to get admission in good quality institute. They must organize very unbreakable in order to be appropriate the exams. The candidates can get ready from previous years solved matter papers.

17 August 2014 at 12:42

(h)

12 September 2014 at 07:15

:-d

14 September 2014 at 12:51

96233+62513=158746

17 September 2014 at 03:37

plz send the elitmus study material to vigneshkumarkandasamy@gmail.com and the cryptographic problem also

2 November 2014 at 00:00

@-)

24 November 2014 at 14:45

plz send the elitmus study material to faizy.wali@gmail.com and the cryptographic problem also

26 November 2014 at 04:00

please send me material for elitmus and the cryptarithematic solved problems please help me....guys

26 November 2014 at 04:01

mail id pallavirpriya@gmail.com guys pls reply soon:-(

27 November 2014 at 02:59

plz send the elitmus study material to jogideepak92@gmail.com and alsocryptographic problem..

27 November 2014 at 10:05

can anyone xplain tha TAKE A CAKE one????why the middle A???
what is that??

1 February 2015 at 22:06

can any one explain me error-hack=crack

1 February 2015 at 22:07

can u explain me error-hack=crack where k=1

7 February 2015 at 04:03

it should be

M Y N N E
+ L O O K S
= S L E E P Y

otherwise, its impossible to solve

10 July 2016 at 07:31

Send study material though this mail ramsanapareddy@gmail.com

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