(68-a)(68-b)(68-c)(68-d)(68-e)=725.wat is the value of a+b+c+d?

• factors of 725= 5,5 29
  so (68-63)=5 ,(68-63)=5,(68-39)=29, (68-67)=1 ,(69-67)=1
  so a+b+c+d= 232
• 9 years agoHelpfull: Yes(28) No(1)
• (68-63)(68-63)(68-39)(68-67)(68-67)=725
  =5*5*29*1*1
  =725
  and a+b+c+d
  =63+63+39+67=232
• 9 years agoHelpfull: Yes(15) No(1)
• it is possible to find sum of all a+b+c+d+e but not possible to find sum of a+b+c+d as different people can take different values of a,b,c,d,e
• 9 years agoHelpfull: Yes(13) No(0)
• 
  (68-a)(68-b)(68-c)(68-d)(68-e)=5*5*29*1*1
  a=63 b=63 c=39 d=68 e=68
  a+b+c+d=232
• 9 years agoHelpfull: Yes(3) No(0)
• @deepak goswami abcde should b distinct
• 9 years agoHelpfull: Yes(2) No(1)
• see 725 is equal to 5*5*29*1*1 however all the digits should be unique so one of the two duplicats is necessarily negative that is 5*-5*29*1*-1
  
• 9 years agoHelpfull: Yes(2) No(0)
• ans is 311
• 9 years agoHelpfull: Yes(1) No(5)
• a+b+c+d=232
• 9 years agoHelpfull: Yes(1) No(0)
• 242 - a=39,b=63,c=67,d=73,e=36
• 9 years agoHelpfull: Yes(1) No(1)
• =5*5*29*1*1(factors)
  in order to get 5 in first place a=63
  same for b ,c,d also
  so a=63,b=63,c=39,d=1,d=1
• 9 years agoHelpfull: Yes(1) No(2)
• if i take e value 39, the sum will change
• 9 years agoHelpfull: Yes(1) No(0)
• dummy question
• 9 years agoHelpfull: Yes(1) No(0)
• how it is possable(a=63)?plz explain.
• 9 years agoHelpfull: Yes(0) No(0)
• 725= 5 * 5 * 29
  (68-63)(68-63)(68-39)(68-67)(68-67)
  63+63+39+67=232
• 9 years agoHelpfull: Yes(0) No(0)
• 63+63+67+67+39=299
• 9 years agoHelpfull: Yes(0) No(0)
• DEEPAK GOSWAMI
  how you took once 67
  its not possible
  say it clear....
• 9 years agoHelpfull: Yes(0) No(0)
• For all the distinct digits...
  1*5*29*-1*-5
  (68-67)*(68-63)*(68-39)*(68-69)*(68-73)
  Thus 67+63+39+69+73=311
• 9 years agoHelpfull: Yes(0) No(0)