eLitmus Cryptarithmetic Problem – 6

February 17, 2024

Cryptarithmetic Questions asked in eLitmus 6

How To Solve Quickly PrepInsta

1.Find the value of G + A + S ? 
(a) 12(b) 13(c) 14(d) 15
2.Find the value of 2S + R?
(a) 9(b) 5(c) 10(d) 12
3.Value of F ? 
(a) 1(b) 2(c) 3(d) 4

It is highly suggested to go through these before solving

  1. Cryptarithmetic Introduction
  2. How to Solve Cryptarithmetic Problems
Row1             G  A  S
Row2          x  F  B  I
Row3        -------------
Row4          F  T  B  I
Row5       S  S  T  B
Row6    S  A  S  F      
Row7    ----------------
Row8    S  R  I  S  T  I

Step 1

  • S x I= _I
  • S x B= _B
  • S x F= _F

As you are getting the same digit after multiplying with S (last digit). This condition is only satisfied when value of S=1. Hence, S=1

Put S=1 and rewrite the problem.

Row1             G  A  1
Row2          x  F  B  I
Row3        -------------
Row4          F  T  B  I 
Row5       1  1  T  B
Row6    1  A  1  F      
Row7    ----------------
Row8    1  R  I  1  T  I

Step 2

Dividing into subproblems –

Row1             G  A  1
Row2          x        B
Row3        -------------
Row4          1  1  T  B

Also, one more thing that we know from here is

  • B + B = T

Lets start hit and trial method –

Now, you have to start hit and trial with the possible value of B={2, 3, 4, 5, 6, 7, 8, 9}

  • Taking B = 2
  • Thus T = 2 + 2 = 4
  • Let us rewrite the problem
Row1             G  A  1
Row2          x        2
Row3        -------------
Row4          1  1  4  2

Now, we can easily predict the value of A = 7

  • Carry from previous step is 0
  • Thus to get 4 at tens digit possible value of A needs to be either 2 or 7
    • 2 x 2 = 4
    • 7 x 2 = 4 (1 carry to next step)
  • 2 value for A is not possible as B already 2
  • Thus, A = 7

Also, clearly G = 5

  • 1 Carry from previous step
  • (G x 2) + = 11
  • Thus, G = 5

Rewriting the whole problem 

Row1             5  7  1
Row2          x  F  2  I
Row3        -------------
Row4          F  4  2  I 
Row5       1  1  4  2 
Row6    1  7  1  F      
Row7    ----------------
Row8    1  R  I  1  4  I

Step 3

Row1             5  7  1
Row2          x        F
Row3        -------------
Row4          1  7  1  F
  • Step 5 x F = 17
  • Now, this is only possible, when F = 3
    • 5 x 3 + 2(carry) = 17

Rewriting the problem again –

Row1             5  7  1
Row2          x  3  2  I
Row3        -------------
Row4          3  4  2  I 
Row5       1  1  4  2 
Row6    1  7  1  3      
Row7    ----------------
Row8    1  R  I  1  4  I

Step 4

Clearly 

  • I = 3 + 1 + 1 + 1 (carry from previous step)
  • I = 6

Thus, R = 8 

  • R = 1 + 7 (no carry from previous step)
Row1             5  7  1
Row2          x  3  2  6
Row3        -------------
Row4          3  4  2  6 
Row5       1  1  4  2 
Row6    1  7  1  3      
Row7    ----------------
Row8    1  8  6  1  4  6

All vaues are now found

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