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.]

You might be interested in this article |

● Cryptarithmetic Problem 11 |

● Cryptarithmetic Problem 09 |

● Cryptarithmetic Tutorial |

● Cryptarithmetic Multiplication Problem |

The way it explained is very helpful to clear concept.. 🙂 thank you very much

Can you please explain me the third answer?

Hi Ishani,

The Pythagorean theorem states that:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

This can be stated in equation form as

a^{2}+b^{2}=c^{2} i.e.

a^2+b^2=c^2

where c is the length of the hypotenuse, and a and b are the lengths of the remaining two sides.

Pythagorean triples are integer values of a, b, c satisfying this equation.

So, you have to check with the values of alphabets, whether it satisfies the above condition or not.

(a) N, P, E

(b) T, P, E

(c) T, H, A

(d) B, N, S

Best Regards,

ElitmusZone

Related Links: https://en.wikipedia.org/wiki/Pythagorean_theorem

Related Links :https://en.wikipedia.org/wiki/Right_triangle

we can also approach like this.

we have following clues

1. E*E=E

2. P*E=E

POSSIBLE SET OF VALUES OF P AND E ARE=

P{6} E{2,4,8,}

P{3,7,9} E{5}

FROM ABOVE 2 SETS YOU CAN ELIMINATE 1 SET BECAUSE

T H E

*E

_______

P I A E AND HERE P CANT BE 6 BECAUSE MAXIMUM 5*9=45 AND MAX REMAINDERS ARE{0,1,2,3,4,5} SO ________ NONE OF THEM WILL GIVE TENTH DIGIT AS 6.

SP PROCEED WITH SECOND SET ::i.e P{3,7,9} E{5}

AS E IS 5 E*ODD NUMBER = 5 AS UNIT DIGIT AND E*EVEN NUMBER = 0 AS UNIT DIGIT .

5 CANT BE UNIT DIGIT IN

(T H E) *N SO VALUE OF I WILL BE 0 FOR SURE AND N MUST BE A EVEN INTEGER SO POSSIBLE VALUES OF N ARE N{2,4,6,8}

NOW PROCEED IN THE SAME DIRECTION WE HAVE ENOUGH CLUES TO SOLVE