If HOW+MUCH=POWER, then value of P+O+W+E+R=?

• 12 is the ans as the leftmost digit would be 1 (carry) and m should be such that on adding a carry from u+h it should change so we choose 9 for m and thus adding (9=m) +1 (carry) o becomes 9+1=10 so o=0.now we would have to assume values for h+w such that there should be no conflict in number assignment we choose h=7 and w=5 adding we get r=12.so r=2.now c+o should be such that when adding 0 which is the value of 0 that we found earlier the number should change from c to e.so let c=3 then e=c+o(0)+1 (carry). which gives e=4.now u+h should be such that we get w which is equal to 5 that we found earlier.so u(=8)+h(=7) gives w=5 so ading values of p+o+w+e+r we get 12
• 10 years agoHelpfull: Yes(16) No(0)
• HOW can be numbered as 814,MUCH can be numbered as 9638 and therefore POWER=10452
  P+O+W+E+R=12
• 10 years agoHelpfull: Yes(8) No(17)
• can anyone xplain clearly?
• 10 years agoHelpfull: Yes(5) No(0)
• o=0 ; p=1 ; r=2 ; c=3 ; e=4 ; w=5 ; h=7 ; u=8 ; m=9
  h7 o0 w5
  +m9 u8 c3 h7
  p1 o0 w5 e4 r2
  
• 10 years agoHelpfull: Yes(4) No(1)
• H O W
  M U C H
  --------------
  P O W E R
  H+U GIVES THE CARRY ,M and carry also generate carry 1,So
  P SHOULD BE 1,
  O SHOULD BE 0,
  M SHOULD BE 9,
  W+H=H+10-----(1) W+H generates carry 1 then
  C+1=E---------(2)
  H+U=W+10-------(3) H+U generates carry 1
  from equation (1) & (3),
  2W=R+U-------(4)
  now reducing the possibility range of variables,we get
  W---->[4,5]--->[5],
  H---->[7,8]-----[7],
  then,
  R----[2],
  U---[8], THEN REMAINING NUMBERS ARE 3,4, & 6 ,
  C----[3],
  E----[4],
  C E H M O P R U W
  3 4 7 9 0 1 2 8 5
  THEN ,
  P+O+W+E+R=1+0+5+4+2 =12.
• 10 years agoHelpfull: Yes(4) No(1)
• how did M get 9?
• 10 years agoHelpfull: Yes(2) No(0)
• m=9(for sum of 4 digit &3 digit =5 digit only when leftmost digit of 4 digit is 9)
  so p=1 &0 =0
  let c=3
  from remaining digit 2,5,6,7,8 we have to select value of h&w so that h+w gives a carry(coz if not so c(3)+o(0)=3 i.e. false)
  so we select h=7 and w=5 which gives u= 8.
  M(9)U(8)C(3)H(7)
  H(7)O(0)W(5)
  P(1)O(0)W(5)E(4)R(2)=12
• 10 years agoHelpfull: Yes(2) No(0)
• those who cant understand how we r getting m as 9 watch this link on youtube
  
  http://www.youtube.com/watch?v=rlwp_LGfHTM
• 10 years agoHelpfull: Yes(2) No(0)
• P+O+W+E+R=(H+O+W).(M+U+C+H)
  comparing this with popular De'Broglie's equation in digital logic.
  SOP and POS forms.
• 10 years agoHelpfull: Yes(1) No(0)
• ans:12
  
  M = 9 bcoz sum of a 3 digit number and 4 digit number only becomes a 5 digit number if the 4 digit number begins with 9. Also that 5 digit number will always begin with 10. so P=1 and O=0
• 10 years agoHelpfull: Yes(1) No(1)
• IF HOW+MUCH=POWER then ANS WILL BE P+O+W+E+R=HOWMUCH...AM I RIGHT?
• 10 years agoHelpfull: Yes(0) No(14)
• how many apti questions will be theere and how much time is concerned for that?
• 10 years agoHelpfull: Yes(0) No(0)
• Let
  M + Sum value should give a carry
  so M = 9
  then O = 0, P = 1
  H + U = should give carry
  W + H = Should give carry
  So take W = 5, H = 7 => W+H=2 with carry 1
  take U = 8 => H + U = 5 with carry 1
  therefore P = 1, O = 0, W = 5, E = 4, R = 2
  P+O+W+E+R = 12
  Answer is 12
• 10 years agoHelpfull: Yes(0) No(0)
• h=7,o=0,w=5
  m=9,u=8,c=3,h=7
  p=1,o=0,w=5,e=4,r=2
  total=12
  
• 10 years agoHelpfull: Yes(0) No(0)
• Thank u Amith
• 10 years agoHelpfull: Yes(0) No(0)
• I m getting:
  HOW=806
  MUCH=9728
  POWER=10634=1+0+6+3+4=14
  Plz tell me is this correct or not??if not then give explaination as soon as psble....
  
• 10 years agoHelpfull: Yes(0) No(3)
• how=46,much=45 power=91 hence p+o+w+e+r=10
• 9 years agoHelpfull: Yes(0) No(1)
• how people are getting e=4.i am unable to get it,plz can anyone explain this?
  
• 9 years agoHelpfull: Yes(0) No(0)