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

• divya (68-a) should be 29 ,so a = 39 , same as b = 63 , c = 63 , d = 67 :
• 9 years agoHelpfull: Yes(3) No(1)
• we know that factors of 725 : 5*5*29*1
  So we need to get (68-a ) = 29 : a = 39
  So we need to get (68-b ) = 5 : a = 63
  So we need to get (68-c ) = 5 : a = 63
  So we need to get (68-d ) = 1: a = 67
  Hence a +b + c + d = 63 + 63 + 67 + 1 = 194
• 9 years agoHelpfull: Yes(1) No(0)
• 725=29*25=29*5*5*1
  hence a=29,b=5,c=5,d=1
  so a+b+c+d=29+5+5+1=40
• 9 years agoHelpfull: Yes(0) No(2)
• a=63 b=63 , c= 63 d=63 e=1/68
  a+b+c+d=252
• 9 years agoHelpfull: Yes(0) No(0)
• 29*5*5=725 . 68-a=29 ; a= 39 68-b=5; b=63 ; similarly c= 63. 68-d=1 ;d=67 e=1/68
  a+b+c+d = 39+63+63+67= 232.
  
• 9 years agoHelpfull: Yes(0) No(0)
• there are multiple answers possible for this question. so go for none of these.
• 9 years agoHelpfull: Yes(0) No(0)