AGE
* OAT
---------
SOAR
HOG-
GOTO--
---------------
GECOIR
i want to know the procedure to solve this type of queqtions.

• 1) AGE * T = SOAR
  2) AGE * A = HOG ( i.e A is a small no.)
  3) AGE * O = GOTO ( here E * O gives O, so O can be either 0 or 5, but O can not be 0 as AGE * 0 = 000 (OOO); this implies O is 5 & E is an odd no.)
  so putting O as 5 we have AGE * 5 = G5T5 (from here we have G * 5 does not generates a carry as A * 5 gives G5 and that's possible only if G=1 (A1E*5=15T5) which implies A=3)
  so now the crypt is like ->
  31E
  x 53T
  --------
  S53R
  H51x
  15T5xx
  ----------
  1EC5IR ( from here we have I = (3+1) = 4 and C = (5+5+5) = )
  
  Again from 31E * 3 = H51 ( i.e. E * 3 = 1 + carry =2 as next 1 * 3 =3 + carry =5 which left us with the fact that E * 3 = 21 so E = 7. and again we have 3 * 3 = 9 == H; one more thing 317 * 5 = 1585 so T=
  now we have
  317
  x 538
  ----------
  S53R
  951x
  1585xx
  ----------
  17C54R
  and now it looks easier
  1) 317 * 8 = 2536 (so S = 2, R = 6)
  
  2) So, the result will be 170546.(and not 170596 i have mistyped before but now I've Edited it) and Thus C= 0 (My Favorite Digit)
  
  1.G
  2.S
  3.A
  4.I
  5.O
  6.R
  7.E
  8.T
  9.H
  
  Hence The Problem Is Solved. QED
• 12 years agoHelpfull: Yes(13) No(1)
• ___317
  ___538
  -------
  __2536
  __951x
  1585xx
  ------
  170546
  
  
• 12 years agoHelpfull: Yes(8) No(4)
• AGE
  * OAT
  ---------
  SOAR
  HOG-
  GOTO--
  --------
  GECOIR
  
  first of all split the question inn 3 parts
  AGE
  T
  ------
  SOAR
  
  AGE
  A
  ----
  HOG
  
  AGE
  O
  ----
  GOTO
  
  AGE * o giving o at unit place
  it makes to possibilites
  1) o=5 E=1,3,7,9
  2) E=6 o=2,4,6
  
  now see the sum
  u can get from there that o + 1 or 2(carry) = E
  so we can easily get o=5 and E=7
  
  than put this values back
  AG7
  5
  ----
  G5T5
  
  u can get G=1 T=8 A=3 from here.
  
  it is not so much tough.
  u can refer elitmuszone.com site to learn this.
  
• 8 years agoHelpfull: Yes(2) No(0)
• please explain the ans clearly i don't understand how 317 comes
• 12 years agoHelpfull: Yes(0) No(2)
• G MAX 1 AND G NOT EQUAL TO 0 SO "G=1"
  SINCE A*A=H HENCE A
• 12 years agoHelpfull: Yes(0) No(3)
• can anyone explain the concept to solve these type of questions?
• 12 years agoHelpfull: Yes(0) No(2)
• bhai koi to kaidey se explain kar do ho kaise rahe hai is type k ques
  I mean to say from where 317 538 came & how G max 1 or min 0 is assumed????
  please explain this in detail...
• 12 years agoHelpfull: Yes(0) No(1)
• http://cryptarithms.awardspace.us/primer.html
  
  try to get concept from here.
  
• 10 years agoHelpfull: Yes(0) No(0)
• Thankyou ABHISHEK for the link.... :-)
• 9 years agoHelpfull: Yes(0) No(0)
• 170546 is the ans.
• 9 years agoHelpfull: Yes(0) No(0)
• 3*O+Carry=O,
  we can have only 3 no (9,4,5).
  (1)
  If O=9,
  then see in 2nd last column O+Carry=E and We can't take any carry from (O+Carry=E ) , otherwise existence of G won't be there. so we reject 9.
  (2)
  If O=5,
  look at A G E * A= HOG
  so A can have value 0,1,2,3.
  we neglect 0 and 1 (if we will take then A G E * A should be equal to A G E * A ).
  (a)
  take A =2
  2 G E * 2= H 5 G
  from multiplication it is clear that each digit gives only 1 digit after multiplication. so, to
  maintain O=5 , G can have only two values (2,7). G can't be 2 (because A has been already taken
  as 2.) now if G =7 then H will also become 5. so we reject 7 too.
  (b).
  take A=3,
  so, 3 G E *3= H 5 G so H =9 [ no carry because it will lead to 4 digit formation.]
  now, O=5, A=3 , H=9 ,
  look at 3 G E * 5= G 5 T 5 , E can have value (1,3,5,7,9).
  u can't 3 , 5 and 9 . u also can't take 1. so E= 7
  look at 3 G 7* 3= 9 5 G . so G =1
  so we have O=5, A=3 , H=9 , E=7 , G=1 , I=4
  by putting these values u will get answer 3 1 7
  * 5 3 8
  2 5 3 8
  9 5 1
  1 5 8 5
  1 7 0 5 4 6
  
• 9 years agoHelpfull: Yes(0) No(0)