# eLitmus Cryptarithmetic Problem – 6

## Cryptarithmetic Questions asked in eLitmus 6

 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

```Row1             G  A  S
Row2          x  F  B  IRow3        -------------
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```