1. | Value of A ? | |||

(a) 3 | (b) 4 | (c) 5 | (d) 6 |

2. | Find the value of W + P + R ? | |||

(a) 16 | (b) 17 | (c) 18 | (d) 19 |

3. | Value of R ? | |||

(a) 6 | (b) 7 | (c) 8 | (d) 9 |

```
W B A
x B P W
C X R F
F X A X
A A C C
A P C A B F
Use Unit Digit Method to solve the problem. The only way to solve this problem is hit and trial approach.
```

```
1. A
2. D
3. D
W=4, B=7, A=3, P=6, C=1, X=8, R=9, F=2, R=9
4 7 3
x 7 6 4
1 8 9 2
2 8 3 8
3 3 1 1
3 6 1 3 7 2
```

W B A x B P W C X R F F X A X A A C C A P C A B F Firstly divide the problem in three parts. As, (1) W B A (2) W B A (3) W B A x W x P x B C X R F F X A X A A C C Now you have to select one from three which has maximum number of clues.i.e. maximum number of variables getting repeated. In this case take, (3) W B A x B A A C C Now you have to start hit and trial with the possible values C i.e. C={1, 2, 3, 4, 5, 6, 7, 8, 9} Value of C cannot be equal to zero) (1) W B A x W0X R F Cryptarithmetic Tutorial Fundamental Rules Point 03 Firstly take C=1 and check further Possible ways of getting 1 as Unit Digit (3) W B A x B A A C C Case 1 C=1 A=1 B=1 Rejected As A=B=C Case 2 C=1 A=7 B=3 Needs to be checked further Case 3 C=1 A=3 B=7 Needs to be checked further Case 4 C=1 A=9 B=9 Rejected As A=B [In Cryptarithmetic, each variable should have unique and distinct values.] Now check for Case 2 and Case 3 only. Case 2 C=1, A=7, B=3 (3) W B A x B A A C C W 3 7 x 3 7 7 1 1 Rejected As even after taking the value of W=9, You will never get 77. W 3 7 x 3 7 7 1 1 Now, you have to check with Case 3 Case 3 C=1, A=3, B=7 W B A x B A A C C W 7 3 x 7 3 3 1 1 Here you can easily predict the value of W=4 Hence W=4, C=1, A=3, B=7 put these values in main problem 4 7 3 x 7 P 4 1 X R F F X 3 X 3 3 1 1 3 P 1 3 7 F Now, you can see 4 7 3 x 4 1 8 9 2 [ 1 X R F] If you compare side by side you will get X=8, R=9 and F=2 Put these values in main problem, and rewrite again 4 7 3 x 7 P 4 1 8 9 2 2 8 3 8 3 3 1 1 3 P 1 3 7 2 Now you can easily predict the value of P=6 Hence, 4 7 3 x 7 6 4 1 8 9 2 2 8 3 8 3 3 1 1 3 6 1 3 7 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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very helpful for elitmus

there is not hint as such OTHER X+C=9 (MAXIMUM CHANCES ALWAYS TO BE 9 IN SUCH CASES) ,10 OR 8

you can proceed wkith hit and trial and start with

W B A

* B C={1,2,3,,,,,9} C CANT BE ZERO

AACC

why c cant be 0

go for hit and trial for A from options that is given and take WBA *B

it would make easy