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, we 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 possible cases.
Firstly, take E=5 and P = {3, 7, 9}
Put E=5 and rewrite the problem again.
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, 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 in comparison to case (1) and Case(2)
(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,
(2) 6 1 5
x 5
3 0 A 5 [ 3 0 7 5]
If you compare side by side then you will get A=7
Put these values in the main problem,
T=6, H=1, E=5, P=3, I=0, A=7
Hence,
6 1 5
x 3 5 N
S N 6 0
3 0 7 5
1 B N 5
S 1 7 7 1 0
Now you can easily solve the problem.
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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why we took i=0?

can any one explain?

[If you multiply to any number with 5, you will only get [0, 5] as their last digit.]

(5 x even =_0 and 5 x odd=_5)

and we have already taken value of E=5. So, only one possibility for the value of I.

Therefore value of I=0.

thank you for explaining how i=0

I can’t able to read solutions.. wat wil do?? Plz anyone help me..!!

Please go through the cryptarithmetic tutorial.

http://www.elitmuszone.com/elitmus/cryptarithmetic-tutorial/

Tutorial of cryptarithmetic is really helpfull. Give time to well understand tutorial after that i’m 100% sure you can easily solve the problem. After the cryptarithmetic tutorial it need to be understand unit digit method then only you can solve any problem of cryptarithmetic.

why not take E as 1,since ExE=(1,5,6).why only (5,6)

if u multiply a 3 digit number to one the multiplication will also be 3 digit number but in the problem multiplication is 4 digit number. ex–THE*E=PIAE; for the value E=1 multiplication will be THE…

You can value of E=1 as well. But later on, E=1 will get rejected.

Please go through the explanation of Rakesh.

Hello Admin… M very bad in cryptography,, as I wrote elitmas bt my PS marks was too bad. Cryptography is one of the better part in ps.. If I will practice these questions available on ur sites,, will it be beneficial ??? 7 Aug is my test date. Kindly help what to do for PS.

Please go through the cryptarithmetic tutorial.

http://www.elitmuszone.com/elitmus/what-is-cryptarithmetic/

Please explain why p={ 3, 7, 9} only

Please go through the Cryptarithmetic Tutorial.

http://www.elitmuszone.com/elitmus/how-to-solve-cryptarithmetic-problems-03/

Rule 3:

If A x B = _ A then possible values of A and B

how can we deduce T=6 by 5*T=30 ?

why are we not considering possibility of remainder there in T*H?

Hi Pranjal.

Query 1:how can we deduce T=6 by 5*T=30 ?

Only one possible value of T exits. 5 x 6 = 30… ( Just write the mulitplication table of 5 in detail and analyse all the possible cases. 🙂

Query 2: why are we not considering possibility of remainder there in T*H?

There is no possibility of Remainder in T * H .

(2) T H 5

x 5

3 0 A 5

You have already taken I=0. If you will get any remained in T * H Multiplication, then value of I cannot be equal to zero. If you have any remaider then value of I can be 1,2,3,4,

Best Regards,

Elitmuszone Team

T = 6

H = 1

E = 5

P = 3

N = 4

S = 2

I = 0

A = 7

B = 8

Can you please clarify after second division of problem into 3 parts which case your considered it should be case 3 but you explained solution using case 2

anybody tell me what are the answer of this cryptarithmetic problem.

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