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

• (68-a)(68-b)(68-c)(68-d)(68-e)=29*5*5*1*1
  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
  a+b+c+d=232
  
• 9 years agoHelpfull: Yes(16) No(3)
• ans is 232 factors of 725
• 9 years agoHelpfull: Yes(1) No(2)
• factors of 725 = 29*5*5*1*1
  so equate ,
  (68-a)=29 similarly (68-5),(68-5),(68-1),(68-1)
  which results in
  a=39,b=63,c=63,d=67,e=67
  a+b+c+d=39+63+63+67=232.
• 9 years agoHelpfull: Yes(1) No(0)
• lcm of 725=5,5,29
  (5)(-5)(1)(-1)(29)=725
  68-a=5:a=63;
  68-b=-5:b=73;
  68-c=1:c=67;
  68-d=-1;d=69;
  68-e=29:e=39;
  a+b+c+d=272
• 9 years agoHelpfull: Yes(0) No(2)
• 725 = 5*5*29*1*1
  so,
  68-a = 5; a = 63
  68-b = 5; b = 63
  68-c= 29; c= 39
  68-d = 1; d = 67
  68-e = 1; e = 67
  
  so, a+b+c+d+e = 299
  and a+b+c+d = 232.
• 9 years agoHelpfull: Yes(0) No(0)
• (68-a)(68-b)(68-c)(68-d)(68-e)=725=29*5*5*1*1=(68-39)(68-63)(68-63)(68-67)(68-67)
  a+b+c+d=39+63+63+67=232
• 9 years agoHelpfull: Yes(0) No(0)