1. | Value of T ? | |||

(a) 3 | (b) 4 | (c) 5 | (d) 6 |

2. | Value of P? | |||

(a) 5 | (b) 1 | (c) 3 | (d) 4 |

3. | Value of K ? | |||

(a) 2 | (b) 8 | (c) 0 | (d) 1 |

```
T P K
x P V A
R 2 E L
T Q W ?
E E V E
? A R T W L
Here, you don't have a single clue to start solving the Cryptarithmetic problem.
You have to use Unit Digit Method (Hit and Trial Approach) for solving this Cryptarithmetic problem.
Now, firstly go through Unit Digit Method and try to solve the problem by your own.
```

```
1. B
2. A
3. B
4 5 8
x 5 9 7
3 2 0 6
4 1 2 2
2 2 9 0
2 7 3 4 2 6
```

(Solution have been given considering you as beginner in Cryptarithmetic) T P K x P V A R 2 E L T Q W ? W W V E ? A R T W L Now you have to divide the problem into three parts. As, (1) T P K x A R 2 E L (2) T P K x V T Q W ? (3) T P K x P W W V E At this stage, you have to analyse all three parts and find which has maximum number of clues.(i.e. maximum number of variables in the repetition) In this problem you can choose (3 OR 1). In this case you can take (3), (1) T P K x P W W V E Now you have to start hit and trial with the possible value of E={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Firstly take E=0 Possible ways of getting 0 at Unit Digit Case 1 - K=5, P=2, E=0 Needs to be checked further Case 2 - K=2, P=5, E=0 Needs to be checked further Case 3- K=4, P=5, E=0 Needs to be checked further Case 4- K=5, P=4, E=0 Needs to be checked further Case 5- K=6, P=5, E=0 Needs to be checked further Case 6- K=5, P=6, E=0 Needs to be checked further Case 7- K=8, P=5, E=0 Needs to be checked further Case 8- K=5, P=8, E=0 Needs to be checked further Firstly take Case 1Case 1 - K=5, P=2, E=0Put these values in (3) (3) T P K x P W W V E i.e. T 2 5 x 2 W W V 0 From here you will get value of V=5. (Rejected, as you have already taken K=5). [In Cryptarithmetic, each variable should have and unique and distinct value.] Cryptarithmetic TutorialCase 2 - K=2, P=5, E=0Put these values in (3) (3) T P K x P W W V E i.e. T 5 2 x 5 W W V 0 From here you will get value of V=6. Now problem reduces to below T 5 2 x 5 W W 6 0 From, here you can assume the possible W =( 1, 3, 4) (You cannot assume the values of W as 0, 2, 5, 6 as you have already taken.) Further you cannot assume the value of W as 7, 8, 9 ) (Please recollect the table of 5. You will never get 6, 7, 8, 9 as last digit whenever you will multiply any number with 5. ) Just take each possible values of W={1, 3, 4} and check with the combination of T. [Relax it's going to take some time to understand the concept. Please read.... again !]Case 3 K=4, P=5, E=0Now put these values in (3) and check further.. (3) T P K x P W W V E T 5 4 x 5 W W V 0 For here you will get the value of V=7. Now you have to search for the possible values of W= {1, 2 , 3} You have to make hit and trial with the possible values of W. Firstly take W=1. now the problem will reduce to (3) T 5 4 x 5 1 1 7 0 [Rejected: For any value of T.. you will never get 11 at last.] Now try with W=2 (3) T 5 4 x 4 2 2 7 0 [Rejected : For any value of T.. you will never get 22 at last..] [Don't think of taking value of T=4. As you have already , E=0 taken K=4]Case 4 K=5, P=4, E=0(3) T P K x P W W V E i.e. T 4 5 x 4 W W V 0 From here you will get value of V=8. Now you have to search for the possible values of W = {1, 2, 3} All the cases will get rejected. Please try yourself for each possible values of W.Case 5 K=6, P=5, E=0(3) T P K x P W W V E T 5 6 x 5 W W V 0 From here you will get value of V=8. T 5 6 x 5 W W 8 0 Now you have to search for the possible values of W= {1, 2, 3, 4} All the cases will get rejected. Please try yourself for each possible values of W.Case 6 K=5, P=6, E=0(3) T P K x P W W V E Put these values in the (3) T 6 5 x 6 W W V 0 From here you will get value of V=9 Now the problem reduces to T 6 5 x 6 W W 9 0 Now just think for all the possible values of W={1, 2, 3, 4} Firstly take W=1 and think of possible values of T. All the cases will get rejected.[ Just try with all combinations]Case 7 K=8, P=5, E=0(3) T P K x P W W V E T 5 8 x 5 W W V 0 From here you will get V=9 Now you have to start hit and trial with the possible values of W={1, 2, 4} Firstly take W=1. For any value of T x 5 + 2 { you will never get equal values W W i.e. 1 1 } [Relax it's going to take sometime to understand the concept. Please read .... again !] now take W=2. For T=4 it will results into 4 5 8 x 5 2 2 9 0 From here you will get T=4, P=5, K=8, V=9, W=2 and E=0. Put these values in the main problem. 4 5 8 x 5 9 A R 2 0 L 4 Q 2 ? 2 2 9 0 ? A R 4 2 L Now you can easily solve the problem further. 4 5 8 x 5 9 7 3 2 0 6 4 1 2 2 2 2 9 0 2 7 3 4 2 6 [If you are aware of unit digit method concepts. You can easily solve this problem.] [Try to solve atleast 15 problems using unit digit method. You can solve any cryptarithmetic problem within 15 mintues. :)] [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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how to prepare for elitmus within two weeks.

https://www.quora.com/What-is-the-best-method-to-prepare-for-the-elitmus-exam-if-you-have-25-days-left

In case 5, why you havn’t take T =4 and W = 2 ?? There is no problem is we take it .

one more way to solve this question.

we have possible values of

p= {5,1,3,4,} and

k = {2,8,0,1}

and if you will look closely you will find value of p cant be 1

and value of k cant be 0 and 1 (according to basic crypt rule)

so only possible set of values for p and k are

p={5,3,4}

k={2,8,1}

possible set of p and k =

{ (5,2),(5,8),(3,2),(3,8),(4,2),(4,8)}

so start hit and trial with these 6 cases and you will get the answer with second element i.e.. p=5 and k=8.

how did u get E=0 in the beginning ?

Please go through the Unit Digit Method Tutorial.

http://www.elitmuszone.com/elitmus/unit-digit-method/