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eLitmus Cryptarithmetic Multiplication Questions

eLitmus Cryptarithemtic Multiplication Problems

Multiplication Probelms Cryptarithmetic for eLitmus

eLitmus Cryptarithmetic problems are generally asked in eLitmus exam and are of high difficulty. You can study most important questions for eLitmus Cryptarithmetic from below –

       J  E
       B  B
       ----
       J  E
    J  E  A
   --------
 B  A  D  E

Assume that E = 4

1.Value of 5B?
(a) 15(b) 10(c) 5
2.Value of B + A?
(a) 7(b) 5(c) 1(d) 8
3.Value of  D ?
(a) 4(b) 3(c) 7(d) Can not be determined
It is highly suggested to go through these before solving
 - Cryptarithmetic Introduction
 - How to Solve Cryptarithmetic Problems
Row 1         J  E
Row 2         B  B
Row 3         ----
Row 4         J  E
Row 5      J  E  A
Row 6     --------
Row 7   B  A  D  E

Assume that E = 4

Step 1

  • Clearly, from the Row 2 row where
  • B is written in common multiplication
  • J E is multiplied with B and results in B, which results in J E.
  • Thus, the B value is 1

Step 2

  • In Row 4 : E + A = E.
  • This, only possible when A value is 0
Row 1         J  E
Row 2         1  1
Row 3         ----
Row 4         J  E
Row 5      J  E  0
Row 6     --------
Row 7   1  0  D  E

Step 3

  • Now in Row 5 and Row 7
  • J + nothing(Row 4) gives us 10.
  • Thus, there must be 1 carry from the previous step and J value must be 9
  • Now, the problem looks like-
Row 1            9  E
Row 2            1  1
Row 3            ----
Row 4  (carry)1  9  E
Row 5         9  E  0
Row 6         --------
Row 7      1  0  D  E

Step 4

  • 9 + E = D
  • Now, from the previous step, there can not be any carry
    • As E + 0 = E, and E can have values between E = {0,9}
    • Max value of E can be 8 as J = 9 (already taken)
  • Thus, step 9 + E = D will have no carry from the previous step
  • However, this is generating carry to the next step
  • Thus, 9 + E = D + 10
  • E – D = 1
  • The only thing that we can conclude is that E and D are consecutive but we can’t deduce their value.

Cryptarithmetic Questions asked in eLitmus 2

The Cryparithmetic methods discussed below are developed by PrepInsta only and available on 2 PrepInsta owned websites only. Anyone copying the method will be legally sued as these are not open source but are PrepInsta’s Proprietary methods  

                H  O  W
             x  C  U  T
             ----------
             D  D  C  P
          D  W  P  W
       D  U  O  A   
       ----------------   
       D  C  E  P  D  P
1.Value of T + O + P ?
(a) 3(b) 8(c) 6(d) 9
2.Value of B ?
(a) 1(b) 2(c) 3(d) 4
3.Value of  2C ?
(a) 13(b) 13(c) 17(d) 11
It is highly suggested to go through these before solving
 - Cryptarithmetic Introduction
 - How to Solve Cryptarithmetic Problems

As, W x U= _ W [ D W P W ] For W the probablities are- Case-1 When W={2, 4, 8} and U={6} For W the probablities are- Case-2 When W={5} and U={3, 7, 9} Taking the first scenerio Take U=6 and problem looks like, H O W x C 6 T ———- D D C P D W P W D 6 O A
—————- D C E P D P

Getting more progress,

              H  O  W 
           x  C  6  T
           ----------
           D  D  C  P
        D  W  P  W
     D  6  O  A   
     ----------------   
     D  C  E  P  D  P

D + 6 + (carry) = C    [Carry may be either 0, 1 or 2]
Hence, problem looks like of C= {7, 8, 9}
Start hit and trail C={7, 8, 9} and W={2, 4 ,8}

C=7 and W=2
when C=7 and W=2 then D=9 [ As, C + W = D] If we take take D=9 then,

             H  O  W
          x  C  6  T
          ----------
          D  D  C  P
       9  W  P  W
    9  6  O  A   
    ----------------   
    0  C  E  P  D  P
D=0 and D=9 two values of D this C=7 and W=2(Not true)
Now, Check the with other possible values of W and C
Let's take C=7 and W=4

then D=1 as C + W = D [last digit]
Taking D=1, C=7 and W=4 now problems looks like
              H  O  4
           x  7  6  T
           ----------
           1  1  7  P
        1  4  P  4
     1  6  O  A   
     ----------------   
     1  7  E  P  1  P
Now, you can easily predict the value of A=8  As, 7 x 4= _ A [last digit]

              H  O  4
           x  7  6  T
           ----------
           1  1  7  P
        1  4  P  4
     1  6  O  8   
     ----------------   
     1  7  E  P  1  P
Now, 
              H  O  4
                 x  7  
           ----------
           1  6  O 8
You can easily predict the value of H=2

              2  O  4
           x  7  6  T
           ----------
           1  1  7  P
        1  4  P  4
     1  6  O  8   
     ----------------   
     1  7  E  P  1  P
Now you can easily predict other values.
Value of T=5 O=3 and P=0


              2  3  4
           x  7  6  5
           ----------
           1  1  7  0
        1  4  0  4
     1  6  3  8   
     ----------------   
     1  7  9  0  1  0
H=2, O=3, W=4, C=7, U=6, T=5, D=1, S=0, E=9

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

Cryptarithmetic Questions asked in eLitmus 3

The Cryparithmetic methods discussed below are developed by PrepInsta only and available on 2 PrepInsta owned websites only. Anyone copying the method will be legally sued as these are not open source but are PrepInsta’s Proprietary methods 

Ques. For the following Cryptarithmetic find the answers to the below questions?

         P  O  P
         M  U  T
        ---------
      S  O  I  U
   G  M  T  U
U  I  R  O
-----------------
U  U  S  M  U  U
1.Value of G + U + T ?
(a) 3(b) 12(c) 6(d) 9
2.Value of S + G ?
(a) 3(b) 9(c) 7(d) 8
3.Value of  2U ?
(a) 4(b) 6(c) 8(d) 2
It is evendent from row 2 that for multiplication U x P = U. So U cannot be 1. Also since,
I + U = U. So I = 0. Now from the 10000th addition, 1 + G + (I)0 = A. So G = 1 and U = 2.
         P  O  P
         M  2  T
        ---------
      S  O  0  2
   1  M  T  2
2  0  R  O
-----------------
2  2  S  M  2  2
Simlarly from 2nd row of mul. 2 x P = 2, Also from first row of T x P = 2.

So P should take 6, and T should take 7. (These are the only possibilities)
If E= 6, then P x T = 6 x 7 = 42. So 4 carry over.  Now 7 x O + 4 = 0. So O = 8.

         6  8  6
         M  2  7
        ---------
      S  8  0  2
   1  M  7  2
2  0  R  8
-----------------
2  2  S  M  2  2

Now S = 4 as 7 x 6 + 6 = 48.  Also M = 3. R =5
This would be the solution

         6  8  6
         3  2  7
        ---------
      S  8  0  2
   1  3  7  2
2  0  5  8
-----------------
2  2  4  3  2  2

G = 1, U = 2, T = 7, R = 5, S = 4, O = 8, E = 6