1. | Value of M ? | |||

(a) 5 | (b) 6 | (c) 7 | (d) 3 |

2. | Value of M + A + T ? | |||

(a) 10 | (b) 11 | (c) 12 | (d) 13 |

3. | Which of the following is the set of even number ? | |||

(a) {Y, A, S} | (b) {Y, T, S} | (c) {Y, R, S} | (d) {Y, E, M} |

```
E Y E
x M A T
S Y I A
G M T A
A I R Y
A A S M A A
As, I + A = A
Therefore, I = 0
Further,
E Y E
x M A T
S Y I A
G M T A
A I R Y
A A S M A A
and E * A =_A [ G M T A]
Two possible cases
Case 1. E=6 and A={2, 4, 8}
Case 2. A=5 and E={3, 7, 9} Detailed Explanation- Rule 3
Now Proceed Further..
```

```
1. D
2. C
3. A
E=6, Y=8, M=3, A=2, T=7, I=0, R=5, S=4
6 8 6
x 3 2 7
4 8 0 2
1 3 7 2
2 0 5 8
2 2 4 3 2 2
```

**(The solution has been given considering you as a beginner in Cryptarithmetic)
**
E Y E
x M A T
S Y I A
G M T A
A I R Y
A A S M A A
As, I + A = A
Therefore, I = 0
Put I=0 and Rewrite the Problem again,
E Y E
x M A T
S Y 0 A
G M T A
A 0 R Y
A A S M A A
and E x A =_A [ G M T A]. Two possible cases
Case 1. E=6 and A={2, 4, 8} or Detailed Explanation- Rule 3
Case 2. A=5 and E={3, 7, 9}
Firstly take Case (1)
E=6, and A={2, 4, 8}
E=6 and I=0 put in the main problem and proceed further,
6 Y 6
x M A T
S Y 0 A
G M T A
A 0 R Y
A A S M A A
Possible value of A={2, 4, 8}
as 6 * A = _A [G M T A]
6 x 2 =_2[12] [considering last digit only]
6 x 4 =_4[24] [considering last digit only]
6 x 8 =_8[48] [considering last digit only]
Now, you have to start hit and trail with the values of A={2, 4, 8}
6 Y 6
x M A T
S Y 0 A
G M T A
A O R Y
A A S M A A
**Case-I **
When you will take A=2 then T=7
as 6 x T=_2[42]
**Case-II **
When you will take A=4 then T=9
As 6 x T=_4[54]
**Case-III**
When you will take A=8 then T=3
As 6 x T=_8[18]
[Relax- It is going to take some time to understand the concept. Please read...again !]
You have to check for each case separately.
Firstly, take Case(I) and proceed further
A=2, T=7
6 Y 6
x 2
G M 7 2 [1 M 7 2]
If you compare side by side, then you will get G=1
you can easily predict the value of Y=8
Now put Y=8
6 8 6
x 2
1 3 7 2 [G M T A]
Therefore M=3, T=7, A=2, G=1
Hence,
6 8 6
x 3 2 7
4 8 0 2
1 3 7 2
2 0 5 8
2 2 4 3 2 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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Hi

I am new to crypt problems , k=8 and y=8 is it possible? or the unique solution we have to find it only for multiplicand and multiplier?

Thank you in advance

Syed

Only unique solution is possible. no two variables can be represented by same number.

Sir when to use Case -1 and when 2 its so complicated to distinguish them….

can you explain ? how to figure it out.

u will hav to think wich case would b r8 for the given ques, by hit and trial method but in d ques. given above u cant use case2 bcz any of them either A or E couldnt get d value 5.

sir your assumptions are always ryt..how come the you know that the we have to take the value of E=6 not A=5?

plz help

i am not getting…

We haven’t made any assumptions while solving the cryptarithmetic problem.

You can use unit method for solving these problems as well.