Cryptarithmetic Problem 2

Solution:

                 V  I  A
              x  G  O  T
              G  R  O  T
           A  A  R  O
        A  I  A  G      
        A  S  T  A  R  T
Here, 
A x T = _T [G R O T]
A x O = _0 [A A R O]
A x G = _G [A I A G]
This is only possible when A=1
Hence, Put A=1 and rewrite the problem again.

               V  I  1
            x  G  O  T
            G  R  O  T
         1  1  R  O
      1  I  1  G      
      1  S  T  1  R  T
At this stage, divide the problem in 3 parts for collecting more clues..
(1)            V  I  1
                  x  T 
            G  R  O  T
(2)            V  I  1
                  x  O
            1  1  R  O
(3)            V  I  1
                  x  G
            1  I  1  G
Now you have to choose one among three based on number of clues.
In this problem, you can take case (3).
(You can also take case (1) and Case (2))
(3)           V  I  1
                 x  G
           1  I  1  G
At this stage you have to start hit and trial with the possible values of G
G = {2, 3, 4, 5, 6, 7, 8, 9}
*G ≠ {1}
(As you have  already taken A=1)
[In Cryptarithmetic, each variable should have and unique and distinct value.]
*G ≠ {0}
(As you are multiplying some number by G in (3). If we take G=0)
then            V  I  1
                    x G
             0  0  0  0 [1  I  1  G]
Now,take G=2
(3)            V  I  1 
                   x 2
            1  I  1  2
*Rejected (you can see 2 x I =_1)

( You will never get unit digit at 1 after multiplying any number by 2.) 
If we multiply any digit by 2 we cannot get last digit as 1.
i.e. 2x1=2, 2x3=6, 2x4=8 2x5=_0, 2x6=_2, 2x8=_6[16] and 2x9=_8[18])

[Relax it's going to take some time to understand the concept. Please read...again!]

Now,take G=3
put G=3 in case(3)
(3)            V  I  1 
                   x 3
            1  I  1  3
Now we can  see I x 3 = _1 [ 1 I 1 3]
(last digit is 1 which is only possible when I=7 (7 x 3 = 21)
Put I=7 in (3)
(3)            V  7  1 
                   x 3
            1  7  1  3
Now, you can easily predict the value of V=5 (3 x 5 = 15 + 2(carry)=17)

(3)           5  7  1 
                 x  3
           1  7  1  3
Therefore, 
V=5, I=7, A=1, G=3, Put these value in main problem and solve further.

              5  7  1
           x  3  O  T
           3  R  O  T
        1  1  R  O
     1  7  1  3      
     1  S  T  1  R  T
Now you can easily predict the other values.
S=8, T=6, R=4, O=2
These values also satisfies the Basic Cryptarithmetic Rules

              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