HAT
*CUP
------
EIUI
EART-
EUPI--
----------
HIEEEI
---------

• 345
  876
  ----
  2070
  2415
  2760
  -----
  302220
  -----
• 11 years agoHelpfull: Yes(14) No(3)
• T*P= number trailing with I
  T*U= number trailing with T
  T*C= number trailing with I
  
  So T=5, I=0;
  
  So P and C are EVEN i.e. any of 2, 4, 6, 8.
  and U is ODD i.e any of 3, 7, 9.
  
  Now see column IRI->E. So R+1= number in format as 1E or a single digit E.
  Now RE, RI, E+U+1(carry generated from column EAP->E)=10. So E+U=9 and from column E->H we get E+1=H. We have derived that U can be any of 7, 9. So for E+U to be 9 U must be
• 11 years agoHelpfull: Yes(7) No(1)
• HAT
  
  * CUP
  
  ------------
  
  EIUI
  
  EART
  
  EUPI
  
  --------------
  
  HIEEEI
  
  ---------------
  
  Explanation:
  
  Looking at the multiplicant HAT the partial multipication T*U gives T which says T=5 and U is odd. Also T with other two values i.e c & p yields is i. Which means i=0 because when odd number multiplied with 5 gives 5 itself which is not the case. Hence c & p are even which gives i=0. Hence the problem looks like this,
  
  HA5
  
  * CUP
  
  -----------------
  
  E0U0
  
  EAR5
  
  EUP0
  
  ------------
  
  H0EEE0
  
  ------------
  
  now looking at the third column from right R+1=E ( Here there is no carry bcz for this column to carry the maximum value should be 9 which makes E=0 which cant be bcz I is already 0) and the second column from right is U+5=E+10. The maximum value of U is >5 and it must be odd for this condition. Which makes values for U as either 7 or 9. If we take U=9 E will be 4. Then the second column from the left which is 4+9+carry=0+1 or 2 for which the carry has to be 7 which is not possible by adding four numbers.Hence U=7 and E=2. Which reduces our problem to,
  
  HA5
  
  * C7P
  
  -----------
  
  2070
  
  2AR5
  
  27P0
  
  -------------------
  
  H02220
  
  --------------------
  
  Now considering R+1=E makes R=1 since E=1. considering second column from left the equation 7+2+carry=0+1or 2. Here the possible carry form prior column could be one so that 7+2+1=10. So H becomes (E+1=H) H=3. Therefore the problem becomes,
  
  3A5
  
  *C7P
  
  -----------
  
  2070
  
  2A15
  
  27P0
  
  --------------
  
  302220
  
  --------------
  
  now considering partial multiplication of 7*3A5=2A15 makes the value of A=4. And considering third column from left i.e 2+A+P=2+10 makes P=6. Hence the problem becomes
  
  345
  
  *C76
  
  -------------
  
  2070
  
  2415
  
  2760
  
  ------------
  
  302220
  
  ------------
  
  Finally to find C we can us partial multiplication of C*345=2760, which makes C=8.
  
  Hence the solution is
  
  345
  
  *876
  
  ------------
  
  2070
  
  2415
  
  2760
  
  ------------
  
  302220
  
  ------------
  
  
  
  Thank you..
  
• 10 years agoHelpfull: Yes(5) No(0)
• can anybdy explain me how to solve such kind of ques.?
  expln wth d hlp of propr basics
  
• 11 years agoHelpfull: Yes(0) No(0)
• Plz explain the method
• 11 years agoHelpfull: Yes(0) No(0)
• Please someone explain such type of questions ? i mean the method of solving?
• 11 years agoHelpfull: Yes(0) No(0)
• 743
  469
  ----
  6687
  4458
  2972
  -----
  348467
• 11 years agoHelpfull: Yes(0) No(8)
• Could anybody please explain in proper manner...?How re you estimating those values?
  
• 11 years agoHelpfull: Yes(0) No(0)
• rajesh paul ur explanation was gud can u explain still more detailed
• 10 years agoHelpfull: Yes(0) No(0)
• 345
  *876
  ----------
  2070
  2415
  2760
  -----------------
  302220
  hint is :- HIEEEI take any value of e from 2-9 (hit and trial) and HAT*U=EART ie T*U=T (T is either 5 and u is odd no(3,7,9) or U is 6 and T is (2,4,8,0)any even no)
• 9 years agoHelpfull: Yes(0) No(0)