(68-a)(68-b)(68-c)(68-d)(68-e)=725
find a+b+c+d

• here 725 can be write as
  725=5*5*29*1*1 so we can take a,b,c,d,e values as 63,63,39,67,67 respectively.
  so ans is a+b+c+d+e=63+63+39+67+67=299..
  for a+b+c+d=232
• 10 years agoHelpfull: Yes(41) No(6)
• (68-a)(68-b)(68-c)(68-d)(68-e)=29*25
  (68-a)(68-b)(68-c)(68-d)(68-e)=29*5*5*1
  (68-a)(68-b)(68-c)(68-d)(68-e)=29*5*1*-1*-5
  (68-a)=29 =>a=39
  (68-b)=5 =>b=63
  (68-c)=1 =>c=67
  (68-d)=-1 =>d=69
  (68-e)=-5 =>e=73
  a+b+c+d=39+63+67+69=238
• 10 years agoHelpfull: Yes(9) No(7)
• ans is 238
  
• 9 years agoHelpfull: Yes(2) No(1)
• 1*2*2.5*5*29=725
  a=67
  b=66
  c=65.5
  d=63
  e=39
  hence, a+b+c+d+e=300.5
• 10 years agoHelpfull: Yes(1) No(1)
• more than 1 ans possible
• 10 years agoHelpfull: Yes(1) No(1)
• So 5×5×29×1×1.....
  63+63+67+67+39=239
  or
  25×29×1×1×1....
  67+67+67+39+43=283
• 9 years agoHelpfull: Yes(1) No(0)
• 5*5*29
  63+63+39+67=232
• 10 years agoHelpfull: Yes(0) No(6)
• first of all factorize 725- >> 29*5*5*1*1
  now try to manipulate all the factors in the form of the equation
  (68-63)*(68-63)*(68-39)*(68-67)*(68-67)
  a =63
  b=63
  c=39
  d=67
  e=67
  now a+b+c+d = 232
• 10 years agoHelpfull: Yes(0) No(0)
• firstly we factorise 725 then we find 5*5*29
  then we put in equation
  (68-63)(68-63)(68-39)(68-67)(68-67)=725
  from where we find the value of
  a=63 b=63 c=39 d=67 e=67
  a+b+c+d=63+63+39+67=232
• 10 years agoHelpfull: Yes(0) No(0)
• please share correct ans
  
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 232
  
  since,
  (68-a)(68-b)(68-c)(68-d)(68-e)=725
  =>(68-a)(68-b)(68-c)(68-d)(68-e)=29*5*5*1
  =>(68-a)(68-b)(68-c)(68-d)(68-e)=29*5*5*1*1
  
  so,
  (68-a)=29 =>a=39
  (68-b)=5 =>b=63
  (68-c)=5 =>c=63
  (68-d)=1 =>d=67
  (68-e)=1 =>e=67
  
  So,
  a+b+c+d = 39+63+63+67 = 232
• 9 years agoHelpfull: Yes(0) No(1)