1. | Value of G ? | |||

(a) 4 | (b) 5 | (c) 6 | (d) 7 |

2. | Value of 3G + 2X ? | |||

(a) 20 | (b) 21 | (c) 22 | (d) 23 |

3. | Which of the following set contains prime number ? | |||

(a) B, R, E | (b) G, A, E | (c) M, R, A | (d) G, M, X |

G A E x M R A X B R X A B K S B R P R G X G X P X Here, you don't have a single clue to start solving this Cryptarithmetic Problem. You have to use Unit Digit Method (Hit and trial approach) for solving this problem. Further you can collect some clues for solving the problem like G A E x M R A X B R X A B K S B R P R G X G X P X i.e. R + S = P and G = B + 1(carry) Firstly go through the Unit Digit Method and try to solve the problem by your own.

```
1. C
2. C
3. A
G=6, A=4, E=3, M=9, R=7, A=4, X=2, B=5, S=1
6 4 3
x 9 7 4
2 5 7 2
4 5 0 1
5 7 8 7
6 2 6 2 8 2
```

(Solution has been given considering you as beginner in Cryptarithmetic) You have to apply Unit Digit Method for solving this Cryptarithmetic Problem. G A E x M R A X B R X A B K S B R P R G X G X P X At this stage you can collect some clues i.e. G A E x M R A X B R X A B K S B R P R G X G X P X R + S = P Now, divide the problem in three parts. (1) G A E x A X B R X (2) G A E x R A B K S (3) G A E x M B R P R At this stage, you have to analyse all three parts and find which has maximum number of clues. In this problem, you can choose Case (1) as it has maximum number of clues. i.e. A and X are repeated two times. Take Case (1) (1) G A E x A X B R X Now, you have to start hit and trial with the possible values of X={1, 2, 3, 4, 5, 6, 7, 8, 9} Possible ways of getting 1 at Unit Digit Case 1 E=1, A=1, X=1 Rejected (E=X=A) Case 2 E=7, A=3, X=1 Needs to be checked Case 3 E=3, A=7, X=1 Needs to be checked Case 4 E=9, A=9, X=1 Rejected (E=A) [In Cryptarithmetic, each variable should have unique and distinct values.] Now you have to only check for Case 2 and Case 3 Case 2 E=7, A=3, X=1 Put these values in Case (1) (1) G 3 7 x 3 1 B R 1 [ X B 1 1] From here you are getting R=1 and you cannot take R=1 as you have already taken X=1. Rejected Case 3 E=3, A=7, X=1 Put these values in Case (1) (1) G 7 3 x 7 1 B R 1 [ 1 B 1 1] Rejected As you have already taken X=1, So value of R cannot be equal to 1 [In Cryptarithmetic each variable should be unique and distinct values.] Now you have to start hit and trial for X=2 Possible ways of getting 2 at Unit Digit Case 1 E=1, A=2, X=2 Rejected (A=X) Case 2 E=2, A=1, X=2 Rejected (E=X) Case 3 E=3, A=4, X=2 Needs to be checked Case 4 E=4, A=3, X=2 Needs to be checked Case 5 E=6, A=2, X=2 Rejected (A=X) Case 6 E=2, A=6, X=2 Rejected (E=X) Case 7 E=8, A=4, X=2 Needs to be checked Case 8 E=4, A=8, X=2 Needs to be checked Case 9 E=7, A=6, X=2 Needs to be checked Case 10 E=6, A=7, X=2 Needs to be checked Case 11 E=8, A=9, X=2 Needs to be checked Case 12 E=9, A=8, X=2 Needs to be checked (1) G A E x A X B R X Put X=2 (1) G A E x A 2 B R 2 Case 3 E=3, A=4, X=2 (1) G 4 3 x 4 2 B R 2 [2 B 7 2] From here you can easily say R=7 (compare side by side) Put R=7, E=3, A=4, X=2 G 4 3 x M 7 4 2 B 7 2 4 B K S B 7 P 7 G 2 G 2 P 2 At this stage, you can easily predict the value of S=1 and M=9 Put these value in the above problem G 4 3 x 9 7 4 2 B 7 2 4 B K 1 B 7 P 7 G 2 G 2 P 2 P=1 Now, you can easily solve the problem further.. 6 4 3 x 9 7 4 2 5 7 2 4 5 0 1 5 7 8 7 6 2 6 2 8 2 [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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5 4 3

*9 7 4