1. | Value of R ? | |||

(a) 1 | (b) 2 | (c) 3 | (d) 4 |

2. | Value of G + R + A + Y ? | |||

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

3. | Which of the following set will form a right angle triangle ? | |||

(a) G, U, E | (b) C, R, Y | (c) G, C, R | (d) D, H, E |

```
C R Y
x G U N
U U R N
E N D U
R A N G
G R A H A N
As,
Y x N = _ N [ U U R N ]
Y x U = _ U [ E N D U ]
Y x G = _ G [ R A N G ]
Hence value of Y=1
Put Y=1 and rewrite the problem again
C R 1
x G U N
U U R N
E N D U
R A N G
G R A H A N
Now proceed further...
```

```
1. B
2. D
3. A
C=9, R=2, Y=1, G=3, U=5, N=6, A=7, D=0, H=8, E=4
9 2 1
x 3 5 6
5 5 2 6
4 6 0 5
2 7 6 3
3 2 7 8 7 6
```

```
C R Y
x G U N
U U R N
E N D U
R A N G
G R A H A N
As,
Y x N = _ N [ U U R N ]
Y x U = _ U [ E N D U ]
Y x G = _ G [ R A N G ]
Hence value of Y=1
Put Y=1 and rewrite the problem again
C R 1
x G U N
U U R N
E N D U
R A N G
G R A H A N
At this stage you can collect some clues
C R Y
x G U N
U U R N
E N D U
R A N G
G R A H A N
R + 1(carry) = G
R + U = A
Now, divide the above problem into three parts
(1) C R 1
x U
U U R N
(2) C R 1
x U
E N D U
(3) C R 1
x G
R A N G
Now you have to analyse all three parts separately.
In Case (1) and Case (2) you don't have much clue to continue.But if you see (3)
You have one relation between R and G i.e
R + 1(carry) = G
Hence you can start hit and trial with the possible values of R={2, 3, 4, 5, 6, 7, 8, 9}
[Value of R cannot be equal to Zero. (0 A N G).Basic Cryptarithmetic Rule Point (3)]
(3) C R Y
x G
R A N G
Case 1 R=2 G=3
Case 2 R=3 G=4
Case 3 R=4 G=5
....
Now take Case 1 and put R=2 and G=3 in (3)
(3) C 2 1
x 3
2 A N 3
Here you can easily predict the value of N=6 and the possible value of
C= {7, 8, 9}[You will only get 2 at thousand place if C= 7, 8, 9]
Now you have to start hit and trial with the possible values of C={7, 8, 9}
[Relax it is going to take some to understand the concept. Please read ... again!]
Firstly take C=7
(3) C 2 1
x 3
2 1 6 3
You cannot take A=1 as you have already taken Y=1
Rejected
Now check for C=8
(3) 8 2 1
x 3
2 4 6 3
Earlier you have one clues i.e. R + U = A
2 + U = 4
To satisfy this relationship value of U should be equal to 2.As you have already taken R=2, you cannot take U=2.
Rejected
[Relax it is going to take some to understand the concept. Please read.... again !]
Now check for C=9
(3) 9 2 1
x 3
2 7 6 3 [ R A N G ]
i.e. R=2, A=7, C=9, G=3 and N=6
put these values in main problem and check further
9 2 1
x 3 U 6
U U 2 6
E 6 D U
2 7 6 3
3 2 7 H 7 6
Now you can easily predict the other values
U=5, E=4, D=0, H=8
Hence,
9 2 1
x 3 5 6
5 5 2 6
4 6 0 5
2 7 6 3
3 2 7 8 7 6
[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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That was smart #elitmuszone.. (Y) liked it…and thanks