1. | Value of A + S + T + A + R + T ? | |||

(a) 21 | (b) 26 | (c) 24 | (d) 25 |

2. | Find the value of 4R + T ? | |||

(a) 19 | (b) 20 | (c) 21 | (d) 22 |

3. | Find the value of 2A + R ? | |||

(a) 5 | (b) 6 | (c) 7 | (d) 8 |

```
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
Now proceed further...
```

1. B 2. D 3. B V=5, I=7, A=1, G=3, O=2, T=6, R=4, I=7, S=8 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

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 I12 *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)57 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 [Relax - It is 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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in hint why cant this be the case when a = 6 and {t,g,o}= {2 ,4 ,8} as given in tutorial

If you understand then please clear it.

Why can’t we take the value of A as 6 and T=2.O=4 and G=8?? Why should we take 1?

A can also be 6 as explained in tutorial.Is it not ?

If you understand then please clear it.

In hint why A=1 is the only possible value of A.

A can be 6 as well, then {T,G,O}= {2 ,4 ,8} can take any values in these three,

Please explain.

i too got the same

Don’t GIve Up!!!!!

571

* 326

————————–

3426

1142

1713

————————–

186146

in this probelm your solution directly assumed as a = 1but why you can also take as a = 6 and then so you can assume as t, o,g, can belong to 2,4,8 as a*t = t, a*o = o, a*g= g implies that a can be 6 and so if t may be one among {2,4,8} so that 6 *2= _2 (last digit), 6*4 = _4 last digit, 6*8=_8 last digit then why cant we assume as a = 6 and solve the problem, you mentioned directly that a= 1 is the only alternative.

we cannot take this value then the value of r is coming same

why to stop for value of G=3 ?? it can be a possible case to take G=7 then I would be 3… isnt that possible?

Guys, I also got the same thought. I guessed that this could help:

Note that in the second column from right the addition goes O+O=R i.e., 2O=R. Also if you’ve got A=6 then G, O, T must be in {2,4,8}.

Now,

2O = R

Then possible values of R can be

2*2=4, or

2*4=8, or

2*8=16 (ending with 6)

which contradicts all the values of G, O, T and A as well.

you are all correct A={6} T,O,G={2,4,8}

But if take sum at G+A+A=T

CASE 1:

A=6 , G=8

8+6+6=T (G+A+A=T)

T=20 (CASE IS WRONG BECAUSE T IS NOT BECOME 0)

CASE 2:

A=6, G=4

4+6+6=T (G+A+A=T)

T=16 (T IS NOT BECOME 6 BECAUSE WE ALSO TAKE A =6)

CASE 3:

A=6 G=2

2+6+6=T (G+A+A=T)

T=14 (TRUE T IS BECOME 4)

WE DECLARE T=4 & G=2 & O=8(BECAUSE 8 ONLY REMAINING IN THAT POSSIBILITIES SHOWN ABOVE)

THEN,

WE GO ANOTHER SUM O+O=R

8+8=16 (R IS NOT BECOMES 6 BECAUSE A=6 WE ALREADY TAKEN)

SO THEY CANNOT TAKE A=6 IN THIS PROBLEM

thanks

But what about the carry….??

G+A+A+carry(0/1/2)=T