1. | Value of K ? | |||

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

2. | Value of M + X + V + M ? | |||

(a) 9 | (b) 10 | (c) 11 | (d) 13 |

3. | Value of W ? | |||

(a) 0 | (b) 1 | (c) 2 | (d) 3 |

```
H M K
x A V E
A N X X
X A V H
M X V W
M A M V W 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 Cryptarithmetic problem.
Further you can collect some clues.
i.e.
H M K
x A V E
A N X X
X A V H
M X V W
M A M V M X
1. X + H = W
2. X + X + 1(carry)= A
Firstly go through Unit Digit Method and try to solve this problem.
```

```
1. C
2. D
3. A
H=8, M=4, K=6, A=5, V=3, E=7, N=9, X=2, W=0
8 4 6
x 5 3 7
5 9 2 2
2 5 3 8
4 2 3 0
4 5 4 3 0 2
```

```
H M K
x A V E
A N X X
X A V H
M X V W
M A M V W X
(You have to use Unit Digit Method for solving this Cryptarithmetic Problem.)
Firstly divide the problem in three parts.
(1) H M K
x E
A N X X
(2) H M K
x V
X A V H
(3) H M K
x A
M X V W
At this stage, you have to analyse all the three parts and find which has maximum number of clues.
In this problem you can choose (1)(X is getting repeated at unit and tens place.)
Take case (1)
(1) H M K
x E
A N X X
Now you have to start hit and trial with the possible values of X = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Firstly take X=1
Possible ways of getting 1 at Unit Digit
Case 1 K=1, E=1, X=1 Rejected (E = K = X)
Case 2 K=7, E=3, X=1 Needs to be checked
Case 3 K=3, E=7, X=1 Needs to be checked
Case 4 K=9, E=9, X=1 Rejected (K = E)
[In Cryptarithmetic, each variable should have unique and distinct values.]
Now check for Case 2 and Case 3
Case 2 K=7, E=3, X=1
Put these values in (1)
(1) H M 7
x 3
A N 1 1
For getting 1 at tens place value of M should be 3. But you have already taken E=3.Hence value of M cannot be equal to 3. Rejected
Case 3 K=3, E=7, X=1
Put these values in (1)
(1) H M 3
x 7
A N 1 1
For getting 1 at tens place value of M should be 7. But you have already taken E=7. Hence value of M cannot be equal to 7. Rejected
[Relax it's going to take some time to understand the concept. Please read... again!]
Possible ways of getting 2 at Unit Digit
Case 1 K=1, E=2, X=2 Rejected (E = X)
Case 2 K=2, E=1, X=2 Rejected (K = X)
Case 3 K=6, E=2, X=2 Rejected (E = X)
Case 4 K=2, E=6, X=2 Rejected (K = X)
Case 5 K=3, E=4, X=2 Needs to be checked
Case 6 K=4, E=3, X=2 Needs to be checked
Case 7 K=6, E=7, X=2 Needs to be checked
Case 8 K=7, E=6, X=2 Needs to be checked
Case 9 K=8, E=4, X=2 Needs to be checked
Case 10 K=4, E=8, X=2 Needs to be checked
Case 11 K=9, E=8, X=2 Needs to be checked
Case 12 K=8, E=9, X=2 Needs to be checked
Now you have to check for Case 5 to Case 12
Case 5 K=3, E=4, X=2
Put these values in (1)
(1) H M 3
x 4
A N 2 2
Any value of M cannot give 2 at tens place.
[If you multiply any number by 4 you will always get a even number.]
[even(4 x any number) + 1(carry)= odd]
Rejected
Case 6 K=4, E=3, X=2
Put these values in (1)
(1) H M 4
x 3
A N 2 2
At this stage possible value of M=7
put M=7, K=4, E=3, X=2 in the main problem
H 7 4
x A V 3
A N 2 2
2 A V H
7 2 V W
7 A 7 V W 2
Here you can easily predict the value of A=5
H M 4
x 5 V 3
5 N 2 2
2 5 V H
7 2 V W
7 5 7 V W 2
In (1) you can see
H M 4
x 3
5 N 2 2
You will never get 5 at thousand place for any value of H.You are multiplying a three digit number by 3]
Rejected
[Relax it is going to take some to understand the concept. Please read... again!]
Case 7 K=6, E=7, X=2
Put these values in (1)
(1) H M 6
x 7
A N 2 2
At this stage you can easily predict the value of M=4
[7 x 4 + 4(carry) =_2]
Hence put M=4, K=6, E=7 and X=2 in main problem
H 4 6
x A V 7
A N 2 2
2 A V H
4 2 V W
4 A 4 V W 2
At this stage you can easily predict the value of A=5
(As you have already taken K=6 and M=4. So, value of A cannot be equal to 4, 6)
put A=5 and rewrite the problem again.
H 4 6
x 5 V 7
5 N 2 2
2 5 V H
4 2 V W
4 5 4 V W 2
Now you can see W=0 H=8 (2 + H = W)
8 4 6
x 5 V 7
5 N 2 2
2 5 V 8
4 2 V 0
4 5 4 V 0 2
Now you can easily predict other values.
8 4 6
x 5 3 7
5 9 2 2
2 5 3 8
4 2 3 0
4 5 4 3 0 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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answer to second question is not correct.. it should be 13

If M=4 , K=6 & E=7, then also Value of X=2

If M=3, K=7 & E=6, then also Value of X=2

then why those cases are not taken and if we take those values, please let us know

v*m=m is a clue to above problem. then why is it stated that there is no clue??? can anyone please help me??

Because you can get a carry from v*k .That’s why you can’t assume v*m = _M i.e (12+1 = 13 )