A B C
x D E
----------
F E C
D E C
----------
H G B C

• 
  A B C
  x D E
  ----------
  F E C
  D E C
  ----------
  H G B C
  In the second partial product we see D x A = D, hence A = 1.
  
  D x C and E x C both end in C, hence C = 5.
  
  D and E must be odd. Since both partial products have only three digits, neither can be 9. This leaves only 3 and 7. In the first partial product E x B is a number of two digits while in the second partial product D x B is a number of only one digit. Thus E is larger than D, so E = 7 and D = 3.
  
  Since D x B has only one digit, B must be 3 or less. The only two possibilities are 0 and 2. B cannot be zero because 7B is a two-digit number. Thus B = 2.
  
  By completing the multiplication, F = 8, E = 7, and G = 6. The answer is 125 x 37 = 4625
• 11 years agoHelpfull: Yes(32) No(6)
• 125x37 = 4625
  
  125
  x37
  -----
  -875
  375-
  ----
  4625
• 11 years agoHelpfull: Yes(10) No(3)
• please elaborate.I mean how 125 and 37 came.thanks
• 11 years agoHelpfull: Yes(7) No(4)
• Here ABCxDE= HGBC
  
  it means BC can be 25 and D and E will be some odd numbers such that D*C and E*C are having C(5) at unit place ). Also B and C are not having value equal to 1. So values of D and E can be 3,7,9 only.
  Further A should be such that its product with E or D should not give 2 digit number. So A can be 1 only. ( 2 is already booked for B)
  If A is 3, then its product with 7 or 9 will give 2 digit value.
  
  so ABC is 125 only and DE can be 37 or 73 only.
• 11 years agoHelpfull: Yes(6) No(8)
• refer to maxey brooke book for mre cryptic multiplication questions
• 11 years agoHelpfull: Yes(5) No(2)
• Please Explain Garima !!!
• 11 years agoHelpfull: Yes(4) No(6)
• http://cryptarithmetic.wordpress.com/2013/01/25/multiplication-problem-1/
  refer this.... i probably think it will be helpful....
  
• 10 years agoHelpfull: Yes(3) No(1)
• 1 2 5
  x 3 7
  -----------
  8 7 5
  3 7 5
  ----------
  4 6 2 5
  
• 11 years agoHelpfull: Yes(2) No(0)
• plz can any one explian the rules for this problem
• 11 years agoHelpfull: Yes(2) No(4)
• can ny1 help me to understand cryptarithmetic. Is der ny site wich cn help me 2 understand it in a better way
• 11 years agoHelpfull: Yes(0) No(2)
• garima plz tel clearly..
  
• 11 years agoHelpfull: Yes(0) No(1)
• Anyone can tell me how to solve dis type of ques?
• 11 years agoHelpfull: Yes(0) No(4)
• how we know that here d*a =1
• 9 years agoHelpfull: Yes(0) No(0)
• FIRST CHECK WHICH DIGIT CAN NOT BE 0=H,D,F,E,A
  THEN CHECK WHICH DIGIT IS NOT 1=H,D,E,C
  STEP 1:
  HERE C*E=C
  I.E C=5 AND E= ODD DIGIT(3,7,9)
  OR
  E=6 AND C=EVEN NUMBER(2,4,8,)
  LET TAKE C=5,E=3 THEN B=8(E+C)BUT IT NOT SATISFY CONDITION ON MULTIPLYING E=5 WHICH IS NOT POSSIBLE
  NOW TAKE C=5,E=7,THAN B=2 SO D IS ODD NUMBER BCZ(D*C=C AND C=5)POSSIBLE VALUE(3,9)
  PUT D=3 THAN YOU GET H=4(H=D+1) AND VALUE OF A(A1>9 SO VALUE OF A=1 BCZ A*E!>9 WHERE E=7) THEN F=8 G=6 YOU GET
• 8 years agoHelpfull: Yes(0) No(0)
• A B C
  x D E
  --------
  F E C
  D E C
  -------
  H G B C
  ===========
  1 2 5
  x 37
  ------
  8 7 5
  3 7 5
  -------
  4 6 2 5
  --------
  A =1 B= 2 C =5 D = 3 E = 7
  H=4 G=6
• 8 years agoHelpfull: Yes(0) No(0)
• a=1, b=2, c=5, d=3, e=7,h=4
• 8 years agoHelpfull: Yes(0) No(0)
• 1 2 5
  x3 7
  -----------------
  8 7 5
  3 7 5x
  -------------------
  4 6 2 5
• 7 years agoHelpfull: Yes(0) No(0)
• http://www.elitmuszone.com/elitmus/cryptarithmetic-problem-1/
• 7 years agoHelpfull: Yes(0) No(0)
• 125
  x37
  --------
  875
  +375
  --------
  4625
  
  A=1
  B=2
  C=5
  D=3
  E=7
  F=8
  G=6
  H=4
• 1 year agoHelpfull: Yes(0) No(0)