can anyone know the solution
TPK
*PVA
-----------
R2EL
TPW?-
WWVE--
-----------
?ARTWL
------------

• 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 1
  Case 1 - K=5, P=2, E=0
  
  Put 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 Tutorial
  
  Case 2 - K=2, P=5, E=0
  
  Put 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=0
  
  Now 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
• 8 years agoHelpfull: Yes(13) No(3)
• 458
  *597
  ------------
  3206
  4122-
  2290--
  -------------
  273426
  --------------
  it is the toughest question of cryptography all tricks fail in it it you need to follow just hit and trail and believe in ur luck
• 8 years agoHelpfull: Yes(12) No(10)
• Hi friends. Whats is the mean by '?' symbol. Do this symbol have the same value or its value may be different at different different places???
• 8 years agoHelpfull: Yes(12) No(0)
• previous question is wrong the right question is this
  TPK
  *PVA
  -----------
  R2EL
  TQW?-
  WWVE--
  -----------
  ?ARTWL
  ------------
• 8 years agoHelpfull: Yes(6) No(0)
• 458 multiply with 5,7 ok
  but 458*9=4122 its wrong
  
  
• 8 years agoHelpfull: Yes(4) No(1)
• Please find the solution
  
  http://www.elitmuszone.com/elitmus/cryptarithmetic-multiplication-31/
• 8 years agoHelpfull: Yes(3) No(0)
• try 458*597
• 8 years agoHelpfull: Yes(2) No(2)
• @karun chaudhary, I think this is not right question. It is not possible to multiply a 3 digit number with any number from 1-9 so that the number comes out will be of 4 digit number and start with first two digits of given 3 digit number. (TPK*V=TPW?). I may be wrong. But reply fast.
• 8 years agoHelpfull: Yes(2) No(3)
• wrong answer
  
• 8 years agoHelpfull: Yes(1) No(1)
• right prince
  
• 8 years agoHelpfull: Yes(1) No(0)
• tell me the right question
• 8 years agoHelpfull: Yes(0) No(0)
• Ans : (a) 16
• 8 years agoHelpfull: Yes(0) No(1)
• ? means we can put any value instead of question mark which can be helpful and satisfy our encrypting value it can we any value
• 8 years agoHelpfull: Yes(0) No(0)
• can anyone help me how to solve these question when ? comes in it!! I don't know the concept when ? comes in these question
• 8 years agoHelpfull: Yes(0) No(0)
• give me a solutions admin.......please nahi to sach sach bata do
• 8 years agoHelpfull: Yes(0) No(0)
• Hi karun can you tell me the way to solve these type of problems other than hit and trial.
• 8 years agoHelpfull: Yes(0) No(1)
• Instead of TPW? it will be TQW?
  
  Since L
• 8 years agoHelpfull: Yes(0) No(1)
• instead of TPW? it will be TQW?
  since L
• 8 years agoHelpfull: Yes(0) No(1)