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Refer the following links for more crypt arithmetic problems:

Link 1

Link 2

Link 3

 

Q. What is a crypt arithmetic problem?

Ans: It is a mathematical problem in which each letter represents a digit (for example, A = 5). 

Our aim is to find the value of each letter. 

No two letters represent the same digit (If A = 5, B cannot be 5).

The problem could be quite challenging but could be made very easy if solved with certain rules, tricks and presence of mind.

Follow these rules:

1. 0 * n = 0

2. 1 * n = n

3. 5 * odd number = (x)5 where x can be 0/1/3/4, 

     like 5 * 1 = 05 (x = 0)

     5 * 3 = 15 (x = 1)

     5 * 7 = 35 (x = 3)

     5 * 9 = 45 (x = 4)

4. even number * 6 = (x)(even number) where x can be 1/2/4

     like 2 * 6 = 12 (x = 1)

     4 * 6 = 24 (x = 2)

     8 * 6 = 48 (x = 4)

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

Hence, Solved!!

+165 -83

Swati 10/6/2013 2:15:23 PM

Thanks!!! It was very useful! 

ashok 12/4/2013 10:48:19 AM

in step2 .....This multiplication give a number which will always be smaller than 5000 so C can be 3 only. how can say that can u please xplain me

sanjay 12/15/2013 12:45:58 PM

you did a great job.

Payal 12/26/2013 4:33:41 PM

@ashok : "F" can have max value 9 and if u multiply F wid 5 for 2nd row u'll get answer around 4500 i.e. for 2nd row u can never get any answer which will start from 7 or 9, that's why we need to take "C" as 3.

 

And thanks placement yogi... :)

Naveen 12/27/2013 9:27:44 AM

Can anybody elaborate this step?

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

How we came to know that the multiplication is always be smaller than 5000??

 

Thanks Naveen

DEEPAK 1/2/2014 3:31:37 PM

@ASHOK-LET US IMAGINE FAC TO BE 987 IN STEP 2 AS THIS CAN ONLY BE THE MAXIMUM NO. THAT WE CAN MULTIPLY NOW MULTIPLY IT WITH 5 THE MAXIMUM NO. CAN NEVER EXCEED 5000 SO WE ARE LEFT WITH ONLY ONE OPTION IE 3..SINCE THE RESULT IS CJFC SO "C" HAS TO BE LESS THAN 5

sudhakar 1/5/2014 3:16:49 PM

useful way to explain...thanks

 

Ram 1/6/2014 11:22:47 PM

any other method for it sir.

Vijay 1/10/2014 8:54:45 PM

thanx

anam 1/13/2014 12:44:58 AM

not able to get it

mahima tomar 1/29/2014 9:17:35 PM

this is something too awesome..thankyou :)

shruti 2/6/2014 6:36:07 PM

im not understandng :((

 

Vinay Garg 2/15/2014 11:53:57 AM


F A C

* 5

--------

C J F 5

see in this step if we will take c as 9 so in the last C is dere so it comes 9xxx something but we are multiplying 5 with only 3 digits and in 3 digit consider max no i.e. 999 and when u will multiply this 999 with 5 u will get less than 5000 ... i.e. 999*5=4995 so that's why we can't take 5,7,9 and here 3 will come...simple... :)

Sona Shetty 2/18/2014 4:50:09 PM

Awesome explanation!!! Thanks a million!

Bikash Yadav 2/22/2014 12:57:33 PM

thankew very much ...

 

kiran kumar 3/2/2014 10:30:44 AM

i agree with vinay garg so that's why we can't take 5,7,9 and here 3 will come...simple... :)

kiran kumar 3/2/2014 10:59:16 AM

why  "A" is consider as non zero

anubhav 3/5/2014 10:18:26 PM

awesome ek dum gazab pura ka pura concept clear ho gya,good job bro .. thanks a ton

kunal 3/7/2014 8:59:30 AM

when we have two options to choose between odd and even concepts at starting(5 * odd number = (x)5 where x can be 0/1/3/4,even number * 6 = (x)(even number) where x can be 1/2/4).with which one we have to start or hit or trial?

kunal kumar deewan 3/7/2014 11:27:03 PM

@kunal, its your choice. you can start with any of them. Its just matter of luck that you started with is the correct one and you don't have to backtrack to the other option 

 

By the way I have written this whole process, so if anyone has any doubt, he/she can ask me.

Zulekha 3/9/2014 7:03:18 PM

Very helpful thanks.

sadhna 6/2/2014 10:00:42 PM

What about 5*5 in rule 3??? x=2 or smthing else...??

krishn kumar gupta 6/8/2014 12:47:13 PM

   HAT

* CUP

---------------------

          E    I   U   I

     E   A   R   T

E  U    P   I

------------------------

H  I     E    E   E    I

-------------------------

Please sir  solve this .....

kasi viswanath 6/13/2014 10:12:34 AM

pls help me i don't understand how to solve a crypt arthimatic problems.

pls say some basic points

smita 6/17/2014 10:55:52 PM

supper explanation... simple and useful concepts . great job ..... thanks a lot

smita panda 6/18/2014 12:20:38 AM

@Krishn - HAT

               *CUP

      ----------------------------

               EIUT

            EART

          EUPI

----------------------------------

          HIEEEI

ANSWER -   345

                   *876

          ---------------------------

                   2070

                 2415

               2760

       ------------------------------

               302220

 

Pramod Maurya 6/23/2014 2:27:56 PM

@krishn kumar gupta

solution for your HAT * CUP will be

    345

  *876

  2070

 2415

2760

302220

 

 

sahib 6/29/2014 10:40:27 AM

@Pramod Maurya

can u plz describe the solution of HAT*CUP in detail...

sahib 6/29/2014 10:41:29 AM

@Pramod Maurya

can u plz describe the solution of HAT*CUP in detail...

p 8/9/2014 4:40:49 PM

diz hat * cuo is wrong

priyanka 8/14/2014 12:59:59 PM

cant understand this step

"This multiplication give a number which will always be smaller than 5000 so C can be 3 only."
How we came to know that the multiplication is always be smaller than 5000??

Kunal Kumar Deewan 8/14/2014 1:09:39 PM

Priyanka,

    F A C

       * 5

  --------

  C J F 5

This multiplication will be largest when we have this case:

    9 9 9

      * 5

 --------

 4 9 9 5

So it will never exceed 5000.

Hope it is clear now :)

yukti 8/22/2014 12:23:40 PM

it will not exceed 5000.. got this.. but then c will be 3. how come???

Hacker 9/11/2014 10:31:05 AM

Its really good explanation .... but  little bit confusion i hve ? 

amresh 9/12/2014 4:32:31 PM

Above mentioned  HAT*CUP  solution  multipicand T is 5 but in row1 T is 0....is it write...?

chandan 9/21/2014 5:39:39 PM

it is not HAT *CUP it is HAT * CAP

345

876

Piyush 9/25/2014 3:08:19 PM

@yukti

As the maximum value of CJF5 is 4995, therefore C will always be less than 4. As C can have only odd value, it can be 3 only.

pankaj kumar 10/1/2014 12:48:54 PM

very helpful.... realy thanks....Kiss

Akansha 10/2/2014 4:01:20 PM

perfectCool

sadhana 10/3/2014 4:15:48 PM

i read all ur reply still cant get how we r selecting c as 3 plz xplain sir

sadhana 10/3/2014 4:18:11 PM

ty so much its grt i got it now :)

Monali 10/7/2014 5:24:37 PM

CP+IS+FUN = TRUE

can anyone plz explain this

Ajay 10/8/2014 1:39:50 PM

nice explanation

shraddha 10/22/2014 10:39:19 PM

leading characters

means guys...please tell me

 

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Cryptarithmetic Problems eLitmus: How to solve Cryptarithmetic problems asked in eLitmus?

We have posted a step by step solution to cryptarithmetic problems asked in eLitmus. Cryptarithmetic puzzles are always asked in eLitmus as there is atleast one such problem where you have to find the numeric value of alphabets.