1. | Value of N ? | |||

(a) 2 | (b) 4 | (c) 3 | (d) 8 |

2. | Value of T + 2E ? | |||

(a) 15 | (b) 17 | (c) 16 | (d) 11 |

3. | Which of the following forms Right Angled Triangle ? | |||

(a) N, P, E | (b) T, P, E | (c) T, H, A | (d) B, N, S |

T H E x P E N S N T I P I A E H B N E S H A A H I Here E x E = _E [ P I A E ] Therefore, possible values of E= {5, 6} Rule 2 As, 5 x 5 =_5 [25] and 6 x 6=_6 [36] (considering last digit only) Here, you have one more clue, T H E x P E N S N T I P I A E H B N E S H A A H I E x P =_ E Hence, Possible values of E and P are as follows, Case I - When E={5} and P= {3, 7, 9} Case II - When P={6} and E= {2, 4, 8 } Rule 3

```
1. B
2. C
3. A
T=6, H=1, E=5, P=3, I=0, A=7, B=8, N=4, S=2
6 1 5
x 3 5 4
2 4 6 0
3 0 7 5
1 8 4 5
2 1 7 7 1 0
```

(The solution has been given considering you as a beginner in Cryptarithmetic)Firstly, you have to divide the problem in three parts, so that it will help you in collecting more clues. (1) T H E x N S N T I (2) T H E x E P I A E (3) T H E x P H B N E (Choose one among the three which has maximum number of clues.) In this case, you can take Case(2) Here, E x E = _E Therefore, possible values of E = {5, 6} Rule 2 As, 5 x 5 = _5 [25] (last digit) 6 x 6 = _6 [36] (last digit) T H E x P E N S N T I P I A E H B N E S H A A H I Further, you have one more clue E x P = _E Hence, possible values of E and P are as follows. Case I - When E={5} and P={3, 7, 9} Case II - When P={6} and E={2, 4, 8} Rule 3 Now, you have to start hit and trial with both the cases. Firstly, take E=5 and P = {3, 7, 9} T H 5 x P 5 N S N T I P I A 5 H B N 5 S H A A H I Further, 5 x N = _I [ S N T I ] [If you multiply 5 to any number, you will only get [0, 5] as their last digit.] (5 x Even=_0 and 5 x Odd=_5) Therefore, value of I=0 Hence, possible value of N = {2, 4, 6, 8} Now, put E=5 and I=0 and write the problem again. T H 5 x P 5 N S N T 0 P 0 A 5 H B N 5 S H A A H 0 Now, again divide the problem in three parts (1) T H 5 x N S N T O (2) T H 5 x 5 P 0 A 5 (3) T H 5 x P H B N 5 Take Case (2) as it has less number of variable. (2) T H 5 x 5 P 0 A 5 Earlier, you have only three possible values of P={3, 7, 9} you have to start hit and trial with the values of P Firstly, take P=3 (2) T H 5 x 5 3 0 A 5 Then T=6 [6 x 5 = 30] T H 5 x P 5 N S N T 0 P 0 A 5 H B N 5 S H A A H 0 as T + 5 = H i.e. 6 + 5=_1 [last digit] Hence H=1, A=7 (2) 6 1 5 x 5 3 0 7 5 Put these values in the main problem, T=6, H=1, E=5, P=3, I=0, A=7 6 1 5 x 3 5 4 2 4 6 0 3 0 7 5 1 8 4 5 2 1 7 7 1 0 [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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The way it explained is very helpful to clear concept.. thank you very much