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 |

```
G A S
x F B I
F T B I
S S T B
S A S F
S R I S T I
As,
S x I= _I [F T B I]
S x B= _B [S S T B]
S x F= _F [S A S F]
As you are getting the same number after multiplying with S (last digit)
Hence, S=1
Put S=1 and rewrite the problem,
G A 1
x F B I
F T B I
1 1 T B
1 A 1 F
1 R I 1 T I
Now proceed further...
Point to Ponder :
(If you are sure about any particular variable, then replace it and rewrite the problem. It will help you in further analysis of the problem.)
```

1. B 2. C 3. C G=5, A=7, S=1, F=3, B=2, I=6, F=3, T=4 5 7 1 x 3 2 6 3 4 2 6 1 1 4 2 1 7 1 3 1 8 6 1 4 6

```
G A S
x F B I
F T B I
S S T B
S A S F
S R I S T I
As,
S x I= _I [F T B I]
S x B= _B [S S T B]
S x F= _F [S A S 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,
G A 1
x F B I
F T B I
1 1 T B
1 A 1 F
1 R I 1 T I
Now,
G A 1
x F B I
F T B I
1 1 T B
1 A 1 F
1 R I 1 T I
Here, You can collect one more clue i.e. B + B = T
Now, At this stage, divide the problem in 3 parts i.e.
(1) G A 1
x I
F T B I
(2) G A 1
x B
1 1 T B
(3) G A 1
X F
1 A 1 F
Now can take (2) (As it has maximum number of clues i.e. you can see 1 is getting repeated thrice )
You can also take (3) and solve the further.
In this case taking (2)
(2) G A 1
x B
1 1 T B
Now, you have to start hit and trial with the possible value of B={2, 3, 4, 5, 6, 7, 8, 9}
Let's take B=2
then T=4 as B+B=T
Put B=2 and T=4 in (2)
(2) G A 1
x 2
1 1 4 2
Here, you can easily predict the value of A=7, and G=5
[Value of A cannot be 2, as it violates the Basic Cryptarithmetic Rules.You have already taken B=2]
Now, put A=7, T=4 and G=5 in main problem,
5 7 1
x F 2 I
F 4 2 I
1 1 4 2
1 7 1 F
1 R I 1 4 I
Now, you can easily predict the R=9, F=3
5 7 1
x 3 2 6
3 4 2 6
1 1 4 2
1 7 1 3
1 8 6 1 4 6
[Relax - it's going to take some time to understand the whole concept. If you are facing any difficulty. Please go through the Cryptarithmetic Tutorial.]
```

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here, why S cant be 6.

since 6*even no = even no.

and then I,B,F can be even nos.

S can’t be 6 because it makes F,B,I even digits (2,4,8) and none of the even digit satisfy the rules when a hit and try is made on each.

Value of R here must be 8 not 9.

you can proceed like this

just hint

S*F=F

S*B=B

S*I=I

POSSIBLE VALUES OF S {6,1}

IS S=6 THEN POSSIBLE VALUES OF F ,B AND I WILL BE{2,4,8} NOT IN THE SAME ORDER

IF YOU WILL LOOK INTO MULTIPLICATION

B + B=T

IF B=2->T=4 NOT POSSIBLE AS F AND I ,ONE OF THEM WILL HAVE 4

IF B=4->T=8 NOT POSSIBLE AS ONE OF TWO F AND I WILL HAVE 8

IF B=8->T=6 NOT POSSIBLE as s=6

SO REJECT S=6…SO S WILL SURELY BE 1

G A S

*B USE UNIT DIGIT METHOD IN THIS B{2,3,4,,,,,,,,9}

——–

S S T B

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very good question