F Y H
E W T
___________
Z E T F
E K Z Z
Y Z T X
___________
Y F Y W E F

• see,FYH*T=ZETF.IT means Y*T=T.THIS HAPPENS ONLY WHEN T=3 AND Y=4(REMEMBER IT.AS I THINK).BCOZ H*T WOULD HAVE GIVEN A CARRY OF 1 AND THIS CARRY WOULD BE ADDED. i,e 4*3+1=13.
  F4H
  EW3
  ZETF
  
  
  now look ABOVE for H. TO GET A CARRY OF 1 FROM H*3,H CAN BE 5 OR 6. NOW 5 WILL CONTRADICT SINCE IT WILL GIVE F=5(AS H=5). SO H=6
  F46
  EW3
  ZE3F
  EKZZ
  4ZTX
  4F4WEF
  NOW 6*3=18.SO B=8
  WRITE ALL THE FOUND DIGITS AGAIN AND SOLVE..FROM HERE ON ITS EASY
  
  
• 10 years agoHelpfull: Yes(16) No(3)
• F=8
  Y=4
  H=6
  E=5
  W 7
  T=3
  Z=2
  K=9
  X=0
  
  I have used hit & Trial, using F Y H * W = E K Z Z. It took around half hour to do it. Any other lead to solve this problem?
• 10 years agoHelpfull: Yes(11) No(0)
• The Answer is
  846
  573*
  ---------
  2538
  5922*
  4230**
  ---------
  484758
• 10 years agoHelpfull: Yes(2) No(0)
• how y*t=t means y=4,t=3 ?
  it could also mean that y=6 and t=2/4/8
  
• 10 years agoHelpfull: Yes(2) No(0)
• FYH*T=ZETF,=>Y*T=T, IN THIS CASE T=3 & Y=4(WHENEVER this type of case happens)
  put H=6,BY HIT & TRIAL , we get F=8,
  now putting these values, we get, X=0,Z=2,E=5,K=9,W=7,
• 10 years agoHelpfull: Yes(0) No(0)
• from where did you get the value of alphabets ? please explain that one.
• 9 years agoHelpfull: Yes(0) No(0)
• for those who can't understand t is not 5 or 6
  fyh*t=zetf in this let's see y*t=t. if t=5 then there should be no carry at tens place from units place so h=1 but h cannot be 1 as units place is h*t=f and if h=1 it becomes 1*t=f which doesn't satisfy.
• 7 years agoHelpfull: Yes(0) No(0)
• Since E+Z should not give carry, we have a possible value of F is 9,8,7,6,5,4,3.
  (1)Let's first consider F=9 and solve.
  FYH
  xT
  ------
  ZETF ..............(1)
  ------
  the product of two no can only be 9 when we take 3*3=9 and 7*7=49.We cannot take these no in consideration because two letters cannot have same no.
  So Fcannot be 9....
  (2)Let's take F=8,and solve.
  2*4=8,2*9=18,3*6=18,4*7=28,6*8=48.
  (a)considering first(2*4=8)
  T=2 and H=8
  putting this in equation (1).
  we get,Y=6,Z=1,E=7.
  FYH
  xW
  ----
  EKZZ .................(2)
  ----
  Z=1 while having H=8(we cannot get such combination) is not possible from equation(2).
  same is the case with H=2,T=8.
  (b)Now,considering second(2*9=18)
  putting this in equation (1) we cannot get any soution from equation(you can check).
  (c)Now,lets consider third(3*6=18)
  ie.T=3,H=6.
  putting this in equation(1) we get,
  Y=4,Z=2,E=5.
  using these results in equation(2) we get....
  W=7,k=9.
  using these all solution in the question we get..
  F=8,Y=4,T=3,H=6,Z=2,E=5,W=7,K=9,X=0 ANS.
• 7 years agoHelpfull: Yes(0) No(0)